Solved Problems

First decompose the force $F_1$ into its components $F_{1x}$ and $F_{2x}$, then apply Newton's second law to the chosen axes as follows \begin{gather*} \Sigma F_x=ma\\ F_{1x}-F_2=ma \\ \Rightarrow F_1\,\cos \theta-F_2=ma\end{gather*} Substituting the numerical values into the above relation gives us \begin{gather*} 100\,\cos 60^\circ-25=5a_x \\\\  \Rightarrow a_x=\frac{50-25}{5}=5\,{\rm m/s^2}\end{gather*} Since in the vertical direction there is no motion, so the vertical acceleration is zero, i.e., $a_y=0$. Draw a free-body diagram and apply the motion law to the vertical direction. 
\begin{gather*} \Sigma F_y=ma_y \\ N-F_{1y}-mg=0 \\ N=mg+F_1\,\sin \theta=5\times 9.8+100\,\sin 60^\circ \end{gather*} Hence, \[\boxed{N=136\quad \rm{N}}\] 

Refer to the following page for more problems in the AP Physics 1 exam

AP physics 1 forces practice problems with MCQs

In the vertical direction, because the mass is neither moving up nor down: \begin{gather*} \Sigma F_y=0 \\ m_1g=N_1\,\cos \theta \\ \Rightarrow\quad  N_1=\frac{m_1g}{\cos \theta}\end{gather*}
Horizontal direction: \begin{gather*} \Sigma F_x=ma \\ m_1a=N_1\,\sin \theta \\ \Longrightarrow a=g\,\tan \theta\end{gather*}

Now apply Newton's second law to the mass $m_2$
\[\mathrm{\Sigma }F_x=ma\Rightarrow \ F-{\mu }_kN_2-N_1\,{\sin \theta \ }=m_2a\] 
\[N_2=N_1\,{\cos \theta \ }+m_2g\] 
\[\therefore F=m_2a+N_1\,{\sin \theta \ }+{\mu }_kN_2\] 

If we substitute $a\ $and $N_2$ from above into the $F$, we get \begin{align*} F &=m_2g\,\tan \theta+N_1\,\sin \theta+(N_1\,\cos \theta+m_2g)\mu_k\\&= m_2g(\mu_k+\tan \theta)+N_1\,(\sin \theta+\mu_k\,\cos \theta)\\ &= m_2g(\mu_k+\tan \theta)+m_1g(\mu_k+\tan \theta) \end{align*} Hence, we have 
\[\Rightarrow F=(\mu_k+\tan \theta)(m_1+m_2)g\] 
Note: in above, $N_1$ is due to Newton's third law of motion.  

Since we want rolling without slipping and there is a circular motion so the static friction provides the centripetal force acting on the car. Apply Newton's 2${}^{nd}$ law for rotational motions to the car as $\mathrm{\Sigma }F_r=ma_r$
\[F_r=f_{s,max}\Rightarrow \ {\mu }_sN=\frac{mv^2}{r}\Rightarrow \ {\mu }_s=\frac{mv^2}{rN}\] 
Since the car move in the flat circle, $N=mg$. Therefore 
\[\therefore \ {\mu }_s=\frac{v^2}{rg}=\frac{{20}^2}{\left(85\right)\left(9.8\right)}=0.48\ \] 
 

First, draw a free-body diagram, decompose the normal force $\vec{N}$ into its components. Then apply Newton's second law of motion in the $x$ and $y$ direction.
(a) $\mathrm{\Sigma }F_y=N\,{\sin 37{}^\circ \ }-mg=0\ $(mass doesn't move in the $y$-direction)
\[N\,{\sin 37{}^\circ \ }=mg\] 
\[m=\frac{N{\sin 37{}^\circ \ }}{g}=\frac{\left(0.98\right)\left(0.6\right)}{9.8}=0.06\mathrm{kg}\] 
(b) Since the mass $M$ is rotated around a circle so there is a centripetal force in the $x$ direction that is provided by the $x$ component of the normal force $\vec{N}$. Recall that the centripetal acceleration is $a_r=v^2/r$, where $r$ is the radius of the circular motion. thus
\[\mathrm{\Sigma }F_x=ma_x\] 
\[N{\cos 37{}^\circ \ }=\frac{mv^2}{r}\mathrm{\to }v^2=\frac{Nr{\cos 37{}^\circ \ }}{m}\] 
Here, as shown in the figure, 
\[r=\frac{D}{2}+d\,{\sin \theta \ }=\frac{0.08}{2}+0.1{\sin 53{}^\circ \ }=0.04+0.08=0.12\mathrm{m}\] 
Therefore $v=\sqrt{\frac{Nr\,{\cos 37{}^\circ \ }}{m}}=\sqrt{\frac{\left(0.98\right)\left(0.12\right)\left(0.8\right)}{0.06}}=1.25\ \mathrm{m/s}$
 

(a) In the circular motions, the centripetal acceleration is given by $a_r=v^2/r$. So
\[a_r=\frac{{\left(40\right)}^2}{200}=8\ \mathrm{m/s\ }\] 
(b) This centripetal acceleration is provided by the static friction therefore, by the second law we have $\mathrm{\Sigma }F_r=ma_r$
\[f=ma_r\Rightarrow \mu N=ma_r\Rightarrow \mu mg=ma_r\] 
\[\Rightarrow \mu =\frac{a_r}{g}=\frac{8}{9.8}=0.816\] 
Since the road is not banked, so the normal force $N$ balances the downward force of gravity i.e. $N=mg$.
(c) In this case, the horizontal component of the normal force $N\,{\sin \theta \ }$ provides the centripetal acceleration so apply Newton's second law in the $x$ and $y$ directions:
\[\mathrm{\Sigma }F_x=ma_r\Rightarrow N\,{\sin \theta \ }=\frac{mv^2}{r}\] 
\[\mathrm{\Sigma }F_y=0\Rightarrow N\,{\cos \theta \ }=mg\] 
By eliminating $N$ from above equations and solving for $\theta $, we get
\[{\tan \theta \ }=\frac{v^2}{rg}\to \theta ={{\tan}^{-1} \left(\frac{v^2}{rg}\right)\ }=39.2{}^\circ \] 

(d) Let the direction of the motion to be parallel to the incline plane as the $x$ axis. We want the parked car does not slide (move) so the static friction force must be balanced with the parallel component of the weight with the inclined plane i.e. $mg\,{\sin \theta \ }$
\[f_{s,max}=mg\,{\sin \theta \ }\to \ \mu_smg\,{\cos \theta \ }=mg\,{\sin \theta \ }\] 
\[\Rightarrow \ \mu_s ={\tan \theta \ }\] 
\[\mu_s ={\tan 39.2{}^\circ \ }=0.816\] 
(e) When an object moves through a fluid such as water or air, the fluid exerts an opposing force, which is called drag force. The power is given by $P=\vec{F}.\vec{v}$. therefore, the power required to overcome the drag force is 
\[P=F_{drag}v\Rightarrow F_{drag}=\frac{P}{v}=\frac{37300}{40}=933\ \mathrm{N}\] 
 

First, from Newton's second law find the acceleration of the mass and then from kinematic relations finds the time required. 
Let the direction of motion of the elevator be positive of the coordinate system.

\begin{gather*} \Sigma F=ma\to F-mg=ma \\\\ \Rightarrow a=\frac{F}{m}-g=\frac{100}{8}-9.8=2.7\,{\rm m/s^2} \end{gather*} Now, using the relevant kinematic equation below, to find the required time \begin{gather*} y=\frac{1}{2}at^2+v_0t+y_i \\\\  \Delta y=\frac{1}{2}at^2+(0)t\\\\ \Rightarrow t=\sqrt{\frac{2\Delta y}{a}}=\sqrt{2\times \frac{30}{2.7}}=4.7\,{\rm s} \end{gather*}

First, for each of the blocks draw a free body diagram and then apply Newton's 2${}^{nd}$ law to them. Since we want block A does not slide on the B, so its acceleration must be same as that of block B i.e. $a_A=a_B=a$.When the block B moves to the left, the block A also moves to the up along the B so the static friction force between them is down along the block B. For block A we have,
\[\mathrm{\Sigma }F_y=0\to N_A{\cos \theta \ }-f_s\,{\sin \theta \ }-m_Ag=0\] 
\[f_{s,max}={\mu }_sN_A\to N_A\,\left({\cos \theta \ }-{\mu }_s\,{\sin \theta \ }\right)=m_Ag\] 
\[\therefore N_A=\frac{m_Ag}{\left({\cos \theta \ }-{\mu }_s\,{\sin \theta \ }\right)}\] 
\[\mathrm{\Sigma }F_x=m_Aa_A\to N_A\,{\sin \theta \ }+f_s\,{\cos \theta \ }=ma\] 
\[\Rightarrow \ ma=N_A\,({\sin \theta \ }+{\mu }_s\,{\cos \theta \ })\] 
\[\therefore a=g\frac{{\sin \theta \ }+{\mu }_s{\cos \theta \ }}{\left({\cos \theta \ }-{\mu }_s{\sin \theta \ }\right)}\ \ ,\ \ *\ \] 
Now write the equations of motion for block B:

\[\mathrm{\Sigma }F_x=m_Ba_B\to \ P-f_s{\cos \theta \ }-N_A{\sin \theta \ }=m_Ba\] 
\[P-N_A\left({\mu }_s\,{\cos \theta \ }+{\sin \theta \ }\right)=m_Ba\] 
\[P-m_A\underbrace{g\frac{{\mu }_s\,{\cos \theta \ }+{\sin \theta \ }}{{\cos \theta \ }-{\mu }_s{\sin \theta \ }}}_{*\ \Rightarrow a}=m_Ba\] 
Since $m_A=m_B=m$ so 
\[P=2ma=2m\ g\frac{{\sin \theta \ }+{\mu }_s{\cos \theta \ }}{\left({\cos \theta \ }-{\mu }_s{\sin \theta \ }\right)}\] 
Now substitute the given data: $m=10\ \mathrm{kg\ ,\ }\theta=40{}^\circ \ ,\mu_s=0.35$, we obtain
\[P=330\ \mathrm{N}\] 
Note: $N_A$ is the normal force acting on the block A and due to the Newton's 3${}^{rd}$ law its reaction acts on the block B. (see figures)
 

The maximum acceleration of the lower block cannot exceed the maximum acceleration that can be given to the block B by the friction force since the two blocks do not move against each other thus applying Newton’s 2nd law to block B gives (note: we want block B does not slip on the P, so we only have the static friction $f_{s\ max}$)
\begin{gather*} f_s\le f_{s\ max}\left({\mu }_sN\right)={\mu }_s\,mg \\\\ \Sigma F_x=ma_x \\\\ \mu_s\,mg=ma_{max}\\\\ a_{max}={\mu }_sg\quad (*) \end{gather*}
Blocks B and P have simple harmonic oscillation. From Hooke's law  
\begin{align*} F&=kx\\\\ (m+M)a&=kx\\\\ \Rightarrow a_{max}&=\frac{k}{M+m}x_{max}\\\\&=\frac{k}{M+m}A\quad (**)\end{align*}
Substituting ${\mu }_sg$ for $a_{max}$,we obtain \[(*),(**)\Rightarrow A=\frac{M+m}{k}{\mu }_sg\] 
Recall that $\omega =2\pi f=\sqrt{\frac{k}{M_{tot}}}$ so $k=4{\pi }^2\left(M+m\right)f^2$. Substituting this into the above expression for $A$, we will have \[A=\frac{M+m}{4{\pi }^2(M+m)f^2}{\mu }_sg=\frac{{\mu }_sg}{4{\pi }^2f^2}\] 

For practicing more on friction problems, refer 
Coefficient of friction force problems and solutions

(a) Use the following geometry to find the angle $\theta $ in the figure $a$ and $\alpha $ in $b$

\[{\cos \theta \ }=\frac{0.5}{0.625} \Rightarrow \theta =37{}^\circ\]
\[{\cos \alpha \ }=\frac{0.292}{0.417}\Rightarrow \alpha =46{}^\circ \]  
(b) Find the tension in the string by applying equilibrium condition $\mathrm{\Sigma }\vec{F}=m\vec{a}$ to the knot where the strings are connected
Decompose $T$ into horizontal and vertical components. Since there is no motion, $a=0$, apply Newton's 2${}^{nd}$ law to the knot.
\[\mathrm{\Sigma }F_x=2T{\sin \theta \ }-mg=0\] 
\[T=\frac{mg}{2\,{\sin \theta \ }}=\frac{0.5\times 9.8}{2{\sin 37{}^\circ \ }}=4.1\mathrm{N}\] 
(c) apply Newton's 2${}^{nd}$ law to the knot in the figure $b$ 
\[\mathrm{\Sigma }F_y=0\to \ T_3\,{\sin \alpha \ }=mg\to T_3=\frac{mg}{{\sin \alpha \ }}=\frac{0.25\times 9.8}{{\sin 46{}^\circ \ }}=3.4\mathrm{N}\] 
\[\mathrm{\Sigma }F_x=0\to \ T_2=T_3\,{\cos \alpha \ }=\frac{mg}{{\sin \alpha \ }}{\cos \alpha \ }=2.4\mathrm{N}\] 
From the symmetry of the problem, $T_1=T_3=3.4\ \mathrm{N}$
 

(a) All of the boxes have the same acceleration. So for each of the three boxes apply Newton's 2nd law separately. Note that because $m_3>m_1$, we can choose the positive direction as clockwise.  
Box $1$: 
\begin{gather*} \Sigma F_{1y}=0 \\ T_1-m_1g=m_1a \\ \Longrightarrow T_1=m_1(a+g)\end{gather*}
Box 2:
\[\Sigma F_x=m_2a \to \quad T_2-T_1=m_2a\] 
Box 3: \begin{gather*} \Sigma F_{y_3}=m_3a\to \ m_3g-T_2=m_3a \\\\ \Rightarrow T_2=m_3(g-a)\end{gather*}
From the above relations we have  
\[\left\{ \begin{array}{rcl} T_1 & = & m_1(a+g) \\ T_2-T_1 & = & m_2a \\ T_2 & = & m_3(g-a) \end{array}\right.\] Therefore, subtracting $T_2$ from $T_1$ and set equal to the middle equation ($m_2a$), we will have \begin{gather*} m_3(g-a)-m_1(g+a)=m_2a\\\\ \Rightarrow \quad a=\frac{m_3-m_1}{m_1+m_2+m_3}g\end{gather*}Substituting the numerical values into the above equation gives us \[a=\frac{2.5-1.5}{1.5+2.5+3.5}(9.8)=1.3\,{\rm m/s^2}\] 
(b) Substitute the above acceleration $a$ into the $T$'s relations:
\begin{gather*} T_2=m_3(g-a)=(2.5)(9.8-1.3)=21\quad \rm{N}\\\\ T_1=m_1(g+a)=1.5(9.8+1.3)=17\quad {\rm N}\end{gather*}

(a) Because $m_1g\,{\sin 30^\circ} > m_2g$ so the $270\,{\rm g}$ block slides down and thus we can choose the positive direction as counter clockwise. Forces acting on the hanging block are gravity and tension of the rope and the forces acts on the sliding block is gravity ($m_1\ g{\sin 30^\circ}$), normal force($N$) and  tension ($T$).

Apply the Newton's second law to $m_1$ and $m_2$.
To $m_1$: 
\begin{gather*} \Sigma F_{along\ ramp}=m_1a \\\\ m_1g\,\sin 30^\circ-T=m_1a \\\\ T=m_1(g\,\sin 30^\circ-a) \, \quad (1) \\\\ \Sigma F_{normal}=0 \\\\ \to N-m_1g\,\cos 30^\circ=0 \end{gather*} To $m_2$: 
\[\mathrm{\Sigma }F=m_2a\] 
\[T-m_2g=m_2a\] 
\[\therefore T=m_2(g+a)\ \ ,\ \ (2)\] 
If we set the two tension equal each other, we obtain 
\[\left\{ \begin{array}{rcl} T & = &m_1 (g\,\sin 30^\circ-a) \\ T & = & m_2(g+a) \end{array} \right. \] Canceling the tension $T$ from both sides, gives \[ m_1(g\,\sin 30^\circ-a)=m_2(g+a)\] Hence, solving  for $a$ in the above and substituting the numerical values into it, we will have \begin{align*} a&=\frac{m_1\,{\sin 30^\circ}-m_2}{m_1+m_2}g\\\\ &=\frac{0.135-0.075}{0.270+0.075}(9.8) \\\\ &=1.7\quad {\rm m/s^2}\end{align*} Using $a=1.7\,{\rm m/s^2}$, one can determine the tension on the rod as below \begin{align*}T&=m_2(g+a) \\ &=0.075(1.7+9.8) \\ &=0.86\rm{N}\end{align*}
(b) See above.
(c) To find the elapsed time, we can use the kinematic relation such as:
\[x=\frac{1}{2}at^2+v_0t+x_0\] 
In this problem $x-x_0=1\mathrm{m}$ and $v_0=0$ so: 
\begin{gather*} 1=\frac{1}{2}(1.7)t^2+0 \\ \Rightarrow t=\sqrt{\frac{2}{1.7}}=1.1\quad {\rm s}\end{gather*}

(a) First, find the acceleration of the block then use $v^2-v^2_0=2a\mathrm{\Delta }x$ to get the distance traveled up the ramp.
Draw a free body diagram as below: 
Balance the normal forces on the ramp: 
\[N-mg\,{\cos \theta}=0\Rightarrow N=mg\,{\cos \theta }\] 
Apply Newton's second law along the ramp:
\[\mathrm{\Sigma }F_x=ma\to \ -mg\,{\sin \theta \ }-f_k=ma\] 
Let the positive direction of motion toward the up of the ramp. (Because the forces along the ramp are in the opposite direction of the motion of the block so a negative sign is inserted!)

From the above conditions, we obtain: 
\begin{gather*} N=mg\,\cos \theta \\ -mg\,\sin \theta -\mu_k N =ma \end{gather*} Thus, solving for $a$ in the above equations, gives \[ a=-g\,({\sin \theta}+\mu_k\,\cos \theta) \]
The negative indicates that the block is going to come to a stop!
\begin{align*} \Delta x&=\frac{v^2-v^2_0}{2a} \\\\ &=\frac{0-7^2}{-2(9.8)(\sin 25^\circ+0.23\,\cos 25^\circ)}\\\\& =4\quad{\rm N}\end{align*}
(b) Suppose the block does not slide back down, $a=0$ so forces on the block must balance.
\begin{gather*} N=mg\,\cos 25^\circ \\\\ mg\,\sin 25^\circ -f_{s,max}=0 \\\\ \to mg\,\sin 25^\circ -\mu_sN=0\end{gather*} 
Substituting $N$ from first equation into the second, solve for $\mu_s$, we obtain \[\boxed{\mu_s=\tan 25^\circ=0.47} \] 
Note: when the block reaches the highest point of the ramp, the parallel component of the weight ($mg\,{\sin \theta}$) and the friction force ($\mu_{k,s}N$) are in opposite directions. Since $mg\,{\sin \theta}$ is always downward of the ramp but friction forces oppose the relative direction of motion. 

For more problems on the coefficient of friction, refer to: 

Coefficient of friction solved problems

(a) If the block does not slide on the bracket, we can treat the block and bracket as one object. $F$ pulling them and gives the acceleration $F=\left(m_5+m_{10}\right)a$. Therefore, the block is accelerating to the right with $a=\frac{F}{m_5+m_{10}}$. Draw a free body diagram for the block and then apply Newton's second law to it.
\begin{gather*} \Sigma F_y=0\Rightarrow N-mg=0\\ \Rightarrow N=m_{10}g \\\\ \Sigma F_x=ma \Rightarrow f_{k,s}-T=m_{10}a\quad (*) \end{gather*}
The tension on the rope must be balanced with external force $F$, otherwise, the rope will break! So $F_{max}=T$.

Note: since we want to block and bracket moves together, so the positive direction of motion has been chosen to the right.

Note: because the$10\,{\rm kg}$ block does not move on the bracket so friction force is static $f_{k,s}=f_{s,max}={\mu }_sN$.
From ($*$), we have \begin{gather*}{\mu }_sN-F_{max}=m_{10}a\\\\ {\mu }_sm_{10}g-F_{max}=m_{10}\frac{F_{max}}{m_{10}+m_5}\\\\ \Rightarrow {\mu }_sm_{10}g=F_{max}\left(1+\frac{m_{10}}{m_{10}+m_5}\right) \end{gather*} Therefore, the maximum force is obtained as below \begin{align*} F_{max}&={\mu }_sm_{10}g\left(\frac{m_5+m_{10}}{m_5+2m_{10}}\right)\\\\&=(0.4)(9.8)\left(\frac{5+10}{5+2(10)}\right) \\\\ &=23.5\quad{\rm N}\end{align*} 
(b) Since the maximum force $F$ applied to the system so both block and bracket have the same acceleration \[a_{max}=\frac{F_{max}}{m_5+m_{10}}=\frac{23.5}{5+10}=1.56\,{\rm m/s^2}\] 

Practice these questions to understand friction force
Problems on coefficient of friction

Mass $m_1$ is moving around a circle on the table, so it must have a centripetal acceleration. Since the only horizontal force on $m_1$ is the tension force $T$ on the string toward the center of the circle so applying Newton's second law in this radial direction gives us \begin{gather*} \Sigma F_{radial}=m_1a_r \\\\ \Rightarrow T=m_1\frac{v^2}{r} \\\\ \text{where} \quad v=r\omega=r\left(\frac{2\pi }{T}\right)\end{gather*} This same string is holding up $m_2$, so balance the vertical forces on $m_2$ (because $m_1$ moves around a circle with constant radius therefore $m_2$ does not move vertically) \begin{gather*} \Sigma F_{vertical}=0 \\\\ T-m_2g=0 \\\\ \Rightarrow \quad T=m_2g \end{gather*} From above results, one can write \begin{gather*} T=\frac{m_1}{r}{\left(\frac{2\pi r}{T}\right)}^2=m_2g \\\\ \Rightarrow r=\frac{m_2gT^2}{4{\pi }^2m_1} \end{gather*}

(a) The free-body diagram is shown below.

(b) Because there is no vertical motion, $\Sigma F_y=0\Rightarrow f_s-mg=0$. Therefore, the static friction force is \[f_s=mg=75\times 9.8=735\,\rm N\]
(c) If you are moving around a circle, you experience a centripetal force equals to $F_r=m\frac{v^2}{r}$ toward the center of the circle. In this problem, this acceleration is provided by normal force on the body, so $N=F_r=m\frac{v^2}{r}$ but the normal force ($N$) is related to the friction by $f_{s,max}={\mu }_sN$ therefore \begin{gather*} \mu_s N=mg \\\\ \to \ \mu_s \left(m\frac{v^2}{r}\right) =mg \\\\ \Rightarrow \quad v^2=\frac{rg}{\mu_s}\end{gather*} Recall that in the circular motions, tangential velocity is written as $v=r\omega =r\left(\frac{2\pi n}{T}\right)$ where $n$ is the total number of rounds. 
Substituting $r\frac{2\pi n}{T}$ for $v$ in the above equation, we obtain: \begin{gather*} \left(r\frac{2\pi n}{T}\right)^2=\frac{rg}{\mu_s}\\\\ \Rightarrow \quad n^2=\frac{gT^2}{4{\pi }^2{\mu }_sr} \\\\ \Rightarrow n=\sqrt{\frac{9.8\times {60}^2}{4{\pi }^2\times 0.55\times 4}}=20.15\,\rm \frac{rev}{min}\end{gather*}

Note: let the positive direction of the motion as clockwise.
Draw a free body diagram for each block and the pulley then apply Newton's 2${}^{nd}$ law to them.
Note: due to the mass of the pulley, the tension on the rod is different along the two segments. 
\[\mathrm{\Sigma }F=m_4a\Rightarrow T_1=m_4a\ \ ,\ \ (1)\] 
\[\mathrm{\Sigma }F_y=m_2a\Rightarrow m_2g-T_2=m_2a\ \ ,\ \ (2)\] 
$T_1$ and $T_2$ exert opposite torques on the pulley (Newton's 2${}^{nd}$ law for rotation) so 
${\tau }_{net}=I\alpha $ where $\vec{\tau }=\vec{r}\times \vec{F}$ is net external torque on the pulley and $I$is the moment of inertia of the pulley so 
\[\mathrm{\Sigma }\tau =I\alpha \mathrm{\Rightarrow }\left(T_2-T_1\right)R=I\alpha =I\frac{a}{R}\] 
\[\Rightarrow T_2-T_1=\left(\frac{1}{2}m_pR^2\right)\frac{a}{R^2}\Rightarrow T_2-T_1=\frac{1}{2}m_pa\ \ ,\ \ (3)\] 
(Recall that the tangential acceleration of a point on a rotating object relates to the angular acceleration via $a=R\alpha $ where $R$ is the radius of the rotation.)
\[\left(1\right),\left(2\right),\left(3\right)\Rightarrow m_2\left(g-a\right)-m_1a=\frac{1}{2}m_pa\ \] 
Solving for $a$, we obtain
\[\therefore a=\frac{m_2}{\frac{1}{2}m_p+m_4+m_2}g=\frac{2}{\frac{1}{2}\left(0.6\right)+2+4}\left(9.8\right)=3.1\frac{\mathrm{m}}{{\mathrm{s}}^{\mathrm{2}}}\] 
 

Apply Newton's second law in the upward direction, $\Sigma F_y=ma$, gives \[T-mg=ma\] Solving for $a$, gives \begin{align*} a&=\frac{T-mg}{m} \\\\ &=\frac{21750-(1500\times 9.8)}{1500}\\\\& =4.7\quad {\rm m/s^2}\end{align*}

Since the object is in equilibrium state, so apply the equilibrium conditions as 
\[\mathrm{\Sigma }\vec F_x=0\ ,\ \ \mathrm{\Sigma }F_y=0\ ,\ \ \mathrm{\Sigma }\vec {\tau }_P=0\] 
Now draw a diagram and show the forces acting on the object.
\[\mathrm{\Sigma }\vec F_y=0\Rightarrow N-mg=0\Rightarrow N=mg\] 
The torque about a point is defined as $\vec{\tau }=\vec{r}\times \vec{F}$, in this problem choose an arbitrary point, say P, and set the torque about it equal zero to satisfy the equilibrium conditions. Let the positive direction be counter clockwise. Therefore, 
\[\mathrm{\Sigma }\vec \tau =0\to {\tau }_{F_N}-{\tau }_{mg}-{\tau }_f=0\] 
\[\mathrm{\Sigma }\vec {\tau }_P=0\Rightarrow l\underbrace{N}_{mg}{\sin \theta \ }-\left(\frac{l}{2}\right)mg{\sin \theta \ }-lf{\sin \left(\frac{\pi }{2}-\theta \right)\ }=0\] 
\[mg\left(\frac{l}{2}\right){\sin \theta \ }=lf{\cos \theta \ }\Rightarrow f=\frac{1}{2}mg{\tan \theta \ }=\frac{1}{2}\left(\frac{356}{2}\right){\tan 30{}^\circ \ }=51.4\] 
In above, we must insert the weight of one board. 
 

(a)

(b) Apply Newton's second law first on the $m_2$ and $m_3$ to find the tensions, then write again Newton's second law for mass $m_1$ along the ramp to find the unknown mass $m_1$. 
\begin{gather*} \Sigma \vec F_3=0 \\ T_2-m_3g=0 \\ \Rightarrow T_2=m_3g=5\times 9.8=49\,{\rm N} \\\\ \Sigma \vec F_2=0 \\ T_1-T_2-m_2g=0\\ \Rightarrow T_1=49+5\times 9.8=98\, {\rm N} \\\\ \Sigma \vec F_1=0 \\ T_1-m_1g{\sin \theta}=0 \\ \Rightarrow m_1=\frac{T_1}{g\,\sin \theta}=\frac{98}{9.8\,\sin 30^\circ}=20\,{\rm kg}\end{gather*}
 

First, find the acceleration of the box with respect to the incline, then substitute it in the kinematic relation $v^2-v^2_0=2ax$ to determine the $x$.

In above we have illustrated the problem and have drawn a free body diagram. Now apply the Newton's second law in the $x$ and $y$ directions i.e. $\Sigma F_y=0\, ,\, \Sigma F_x=ma$ \begin{gather*} \Sigma F_y=N-mg\,{\cos \theta}=0 \\ \Rightarrow\, N=mg\,\cos \theta \\\\ \Sigma F_x=mg\,{\sin \theta}-\underbrace{f_k}_{\mu_kN}=ma \\ \Rightarrow \, mg\,\sin \theta-\mu_k mg\,\cos\theta=ma\end{gather*}
From the second line, we get
\begin{align*} a&=g\,(\sin\theta-\mu_k\,\cos\theta) \\\\ &=(9.8)(\sin 12.1^\circ-0.360\,\cos 12.1^\circ) \\\\& =-1.395\quad {\rm m/s^2}\end{align*}
The minus sign indicates that the box is going to stop.
\begin{gather*} v^2-v^2_0=2ax \\ 0-(1.5)^2=2(-1.395)x \\ \Rightarrow x=0.806\,{\rm m=80.6\, cm}\end{gather*}

First draw a free body diagram as below on the car
As shown in the figure, two forces acting on the car. The normal force and gravity. Since there is a circular motion, so we have a centripetal force that provided by one of the components of normal force that points to the center of the curved section. Apply Newton's second law as
\begin{gather*} \Sigma F_y=0 \Rightarrow  N\cos \theta=W=mg \\\\ \Sigma F_r=ma_r\Rightarrow N\sin \theta=\frac{mv^2}{r} \end{gather*}
Dividing both equation by each other, we obtain the desired angle $\theta$
\begin{gather*} \frac{N\,\sin \theta}{N\,\cos \theta}=\frac{\frac{mv^2}{r}}{mg}\Rightarrow \tan \theta=\frac{v^2}{rg} \\\\ \Rightarrow \theta ={\tan}^{-1} \left(\frac{(29)^2}{146\times 9.8}\right)=30.44^\circ \end{gather*} 

(a) Two forces, electric and gravity, act on the droplet as shown in the free-body diagram below.

Recall that the force exerted by an uniform electric field $E$ on the charge $q$ is given by ${\vec{F}}_e=q\vec{E}$. Now apply Newton's 2${}^{nd}$ law to this droplet(let upward direction to be $+y$)
\begin{gather*} \Sigma \vec F_y=ma \\\\ F_e\hat{j}+mg(-\hat{j})=ma(-\hat{j}) \\\\ \Rightarrow mg-qE=ma \\\\ \Rightarrow \boxed{m=\frac{qE}{g-a}} \end{gather*} After substituting the numerical values, we have \begin{align*} m&=\frac{(6.24\times {10}^{-18})(9.42\times {10}^3)}{9.8-1.66} \\\\ &=7.22\times {10}^{-15}\, \rm kg \end{align*}
(b) Use the kinematic equation below to find the desired time \begin{gather*} y=\frac{1}{2}at^2+v_0t+y_0 \\\\ -4.24=\frac{1}{2}(-1.66)t^2 \\\\ \Rightarrow t=\sqrt{\frac{2\times 4.24}{1.66}}=2.66\,\rm s \end{gather*} 
Note: we have chosen the reference point as the starting point so $y_0=0$ and since the droplet moving downward, so $y<0$. In the equation above, the acceleration $a$ is a vector quantity. 
(c) To levitate the droplet, all forces acting on it must be in balanced so 
\begin{gather*} \Sigma \vec F_y=0\Rightarrow F_e=mg \\\\ \Rightarrow qE=mg \Rightarrow \boxed{E=\frac{mg}{q}} \end{gather*} Substituting the numerical values, we will get \[E=\frac{(7.22\times {10}^{-15})(9.8)}{6.24\times {10}^{-18}}=1.13\times {10}^4 \,\rm N/C\]

(a) This configuration is called the Atwood's machine. We try to find its tension and acceleration. For this purpose, apply the 2${}^{nd}$ law of motion to each of the masses as
\[\mathrm{\Sigma }{\vec{F}}_{1y}=m_1\vec{a}\Rightarrow T\hat{j}+m_1g\left(-\hat{j}\right)=m_1a\hat{j}\] 
\[\Rightarrow T=m_1\left(a+g\right)\] 
\[\mathrm{\Sigma }{\vec{F}}_{2y}=m_2\vec{a}\Rightarrow T\hat{j}+m_2g\left(-\hat{j}\right)=m_2a\left(-\hat{j}\right)\] 
\[\Rightarrow T=m_2\left(g-a\right)\] 
Equal these two relations to find the acceleration of the system
\[m_1\left(a+g\right)=m_2\left(g-a\right)\Rightarrow a=\frac{m_2-m_1}{m_1+m_2}g\] 
Now substitute $a$ into one of the $T$'s 
\[T=m_2\left(g-a\right)=m_2\left(g-\frac{m_2-m_1}{m_1+m_2}g\right)=\frac{2m_1m_2}{m_1+m_2}g\] 
Therefore, $a=\frac{0.6-0.5}{0.6+0.5}\left(9.8\right)=0.89\ \mathrm{m/}{\mathrm{s}}^{\mathrm{2}}$ and $T=5.3\ \ \mathrm{N}$
(b) By definition, the center of mass of a system of particles with masses $m_i$ is 
\[M{\vec{r}}_{cm}=m_1{\vec{r}}_1+m_2{\vec{r}}_2+\dots \] 
Where $M$ is the total mass of the system and ${\vec{r}}_i$ is position vector of the $i$th particle. To find the CM of a system first establish a coordinate system. For this problem, draw a coordinate that passes through the center of the pulley, as shown: 
Since at the instant of releasing the vertical components of each masses is zero, so $y_{CM}=0$. Now calculate the $x_{CM}$
\begin{align*} x_{CM}&=\frac{m_1x_1+m_2x_2}{m_1+m_2} \\\\ &=\frac{0.5(-25)+0.6(+25)}{0.5+0.6} \\\\ \Rightarrow \quad x_{CM} &=2.77\,\rm mm \end{align*} Therefore, the center of mass of this Atwood's machine is located at $(2.77\,\rm mm ,0)$. 
(c) After releasing, because there is no external force on the machine (such as friction) so the horizontal component of the CM is constant and is the same as previous i.e. $x'_{CM}=2.77\,\rm mm$. To find the vertical component, use the kinematic equation $y=\frac{1}{2}at^2+v_0t+y_0$ as follows \begin{align*} y_{CM}&=\frac{m_1y_1+m_2y_2}{m_1+m_2}\\\\ &=\frac{m_1\left(\frac{1}{2}at^2\right)+m_2\left(\frac{1}{2}\left(-a\right)t^2\right)}{m_1+m_2} \\\\ &=\frac{m_1-m_2}{m_1+m_2}\left(\frac{1}{2}at^2\right) \\\\ &=-\frac{1}{2}\left(\frac{m_2-m_1}{m_1+m_2}\right)^2gt^2 \end{align*} 
By comparison with $y_{CM}=\frac{1}{2}a_{CM}t^2$, we can obtain the acceleration of the CM as below \[a_{CM}=-\left(\frac{m_2-m_1}{m_1+m_2}\right)^2g\] 
Put the given values of masses to determine the $y_{CM}$ and $a_{CM}$
\begin{gather*} y_{CM}=-\frac{1}{2}\left(\frac{0.6-0.5}{0.5+0.6}\right)^2(9.8)t^2=-0.04\,t^2 \\\\ \text{and} \quad a_{CM}=-0.081\,\rm m/s^2 \end{gather*} So the acceleration of the CM is downward.

Apply Newton's 2${}^{nd}$ law to each of the masses. 
Mass $m_1$:
\[\mathrm{\Sigma }{\vec{F}}_y=m\vec{a}\to T\hat{j}+m_1g\left(-\hat{j}\right)=m_1a\left(-\hat{j}\right)\] 
\[\Rightarrow T=m_1\left(g-a\right),\ \ *\] 
Mass $m_2$: 
\[\mathrm{\Sigma }{\vec{F}}_x=m\vec{a}\to T=m_2a\ ,\ \ \ \ \ \ \ **\] 
Now set the tensions $T\ $equal to each other to find the $a\ $and then calculate the tension in the rope.
\[*\ ,\ **\ \ \Rightarrow \ m_1\left(g-a\right)=m_2a\Rightarrow a=\frac{m_1}{m_1+m_2}g\ \] 
Now substitute $a$ into one of the tension relations above:
\[T=\frac{m_1m_2}{m_1+m_2}g=\frac{1\times 1}{1+1}\left(9.8\right)=4.9\frac{\mathrm{m}}{{\mathrm{s}}^{\mathrm{2}}}\] 
 

(a) Since the riders are moving around a circle, so they experience a centripetal force. This force is always points toward the center of the circle i.e. $-\hat{r}$. The two forces, gravity and normal force, acting on the riders are provided the centripetal acceleration as $\mathrm{\Sigma }F_r=ma_r=mr{\omega }^2$. 
\[N\left(-\hat{r}\right)+mg\left(-\hat{r}\right)=mr{\omega }^2\left(-\hat{r}\right)\Rightarrow N=m\left(r{\omega }^2-g\right)\] 
\[=60\left(5\times {\left(1.6\right)}^2-9.8\right)\] 
\[\therefore \ N=180\ \mathrm{N}\] 
(b) At the bottom of the loop, again, the normal force is points towards the center and instead gravity is away from the center. (as shown in the figure)
\[\mathrm{\Sigma }{\vec{F}}_r=m{\vec{a}}_r\to N\left(-\hat{r}\right)+mg\hat{r}=mr{\omega }^2\left(-\hat{r}\right)\Rightarrow N=m\left(g+r{\omega }^2\right)\] 
\[\Rightarrow N=60\left(9.8+0.5\times {\left(1.6\right)}^2\right)=1360\ \mathrm{N}\] 
 

(a) As shown in the free body diagram of A, three forces act on the mass A. if we suppose that at the instant the block A is moving, the block B is motionless, then the frictional force acting on A, due to the B is in the opposite direction of $F$. So 
\[\mathrm{\Sigma }\vec F_A=0\ \left(constant\ speed\ ,a=0\right)\] 
\[\Rightarrow F-f_{k_1}-f_{k_2}-T=0\Rightarrow F=f_{k_1}+f_{k_2}+T\] 
Now apply Newton's 2${}^{nd}$ law to the mass B. Since we have supposed it is motionless so the tension in the rope is equal to the friction force between A and B. 
\[\mathrm{\Sigma }\vec F_B=0\Rightarrow T-f_{k_1}=0\Rightarrow T=\mu_{k_2}m_{B}^{\vphantom{\ast}}g\] 
Substituting $T$ and frictional forces in the relation of $F$, we obtain
\[F=\mu_{k_1}^{\vphantom{\ast}}\left(m_A^{\vphantom{\ast}}+m_B^{\vphantom{\ast}}\right)g+\mu_{k_2}^{\vphantom{\ast}}m_{B}^{\vphantom{\ast}}g+\mu_{k_2}^{\vphantom{\ast}}m_{B}^{\vphantom{\ast}}g\] 
\[\Rightarrow F=\underbrace{\left(0.25\right)\left(5+20\right)\left(9.8\right)}_{f_{k_1}=61.25\ \mathrm{N}}+2\underbrace{\left(0.15\right)\left(5\right)\left(9.8\right)}_{f_{k_2}=7.35\ \mathrm{N}}=76\ \mathrm{N}\] 
(b) In general case, the system has an acceleration. So in this case first find it and then calculate $F_{max}$. Repeat the steps above but here $\mathrm{\Sigma }F=ma$. Therefore
\[\mathrm{\Sigma }\vec F_B=m_B^{\vphantom{\ast}}a\Rightarrow T-f_{k_1}=m_B^{\vphantom{\ast}}a\Rightarrow T={\mu }_{k_2}m_B^{\vphantom{\ast}}g+m_B^{\vphantom{\ast}}a\] 
It has been said the maximum tension in the string is $T_{max}=100\ \mathrm{N}$ so find the acceleration associated with this tension as
\[a_{max}=\frac{T_{max}-\mu_{k_2}^{\vphantom{\ast}}m_{B}^{\vphantom{\ast}}g}{m_B}=\frac{100-\left(0.15\right)\left(5\right)\left(9.8\right)}{5}=18.53\ \mathrm{m/}{\mathrm{s}}^{\mathrm{2}}\] 
Now apply the 2${}^{nd}$ law to the mass A and substitute $a_{max}$ into it to find the $F_{max}$
\[\mathrm{\Sigma }F_A=m_Aa_{max}\Rightarrow \ F_{max}-f_{k_1}-f_{k_2}-T_{max}=m_Aa_{max\ }\] 
\[\Rightarrow \ F_{max}=100+61.25+7.35+\left(20\right)\left(18.53\right)=539.2\ \mathrm{N}\] 

(a) The friction force produces torque on the cylinder. Recall that the formula for the torque and moment of inertia are $\vec{\tau }=\vec{r}\times \vec{F}$ and $\tau =I\alpha$, respectively, where $\alpha$ is the angular acceleration that relates to that of linear via $a=\alpha r$
\begin{gather*} rF=I\alpha \Rightarrow rf_s=\frac{Ia}{r}\\\\ \to r(\mu_s mg\,{\cos 30^\circ})=\frac{\left(\frac{1}{2}mr^2\right)a}{r}\\\\ \Rightarrow \quad \mu_s=\frac{a}{2(g\cos 30^\circ)}\end{gather*}
Now apply Newton's second law to find the linear acceleration of the cylinder, then substitute it into the relation above.
\begin{gather*} \Sigma \vec F=m\vec a \\\\ \Rightarrow mg\,{\sin 30^\circ}-\underbrace{\mu_k mg\,{\cos 30^\circ}}_{f_s}=ma\\\\ \Rightarrow a=g({\sin 30^\circ}-\mu_s{\cos 30^\circ}) \end{gather*} Therefore, the coefficient of static friction gets as \begin{align*} \mu_s&=\frac{g\,({\sin 30^\circ}-{\mu }_s\,{\cos 30^\circ})}{2(g\,{\cos 30^\circ})}\\\\ &=\frac{1}{2}\left({\tan 30^\circ}-{\mu }_s\right) \\\\ \Rightarrow \mu_s&=\frac{1}{3}{\tan 30^\circ}\\\\& =0.192 \end{align*}
(b) With the given $a$, use the kinematic relation $y=\frac{1}{2}at^2+v_0t+y_0$ to find $t$. ($\Delta y=y-y_0$ is the displacement on the incline that is equal to $\Delta y=\frac{2}{\sin 30^\circ}=4\,{\rm m}$)
\begin{align*} a&=g\left({\sin 30^\circ}-{\mu }_k{\cos 30^\circ}\right)\\ &=(9.8)\left(0.5-(0.192){\cos 30^\circ}\right)\\ &=3.27\quad {\rm m/s^2} \end{align*} Hence, the displacement is \begin{align*} \Delta y&=\frac{1}{2}at^2+0\\\\ \Rightarrow \quad t&=\sqrt{2\frac{\Delta y}{a}}\\\\ &=\sqrt{\frac{2(4)}{3.27}}\\\\ &=1.56\quad {\rm s}\end{align*}
(c) In rotational motions, the angular speed is related to the angular acceleration by \begin{align*} \omega=\alpha t&=\left(\frac{a}{r}\right)t\\\\ &=\left(\frac{3.27}{0.5}\right)(1.56) \\\\ &=10.24\quad {\rm \frac{rad}{s}}\end{align*}

To practice more problems on coefficient of friction, refer to: 
Problems on coefficient of friction force

(a) Draw a free body diagram in the general case as shown. Apply Newton's 2${}^{nd}$ law in central and tangential directions to rope. 
\[\mathrm{\Sigma }\vec{F}=m\vec{a}\]

\[\to \ \left\{ \begin{array}{rcl}
T\left(-\hat{r}\right)+W\,{\cos \theta \ }\left(\hat{r}\right) & = & ma_r\left(-\hat{r}\right)\Rightarrow T =W{\cos \theta \ }+ma_r \\ 
W\,{\sin \theta \ } & = & ma_x \end{array}
\right.\] 
Since ${\cos \theta \ }$ in the interval $[0,\pi /2]$ decreases, so if we set $\theta =0{}^\circ $ one can get the maximum value of the tension i.e. $T_{max}=W+ma_r$.
(b) Now use the conservation of energy between $55{}^\circ $ and the angle $\theta $ at which the tension reaches its maximum and breaks. (we have chosen the lowest point of the swing as the base)
\[E_A=E_B\to \ \ K_A+U_A=K_B+U_B\] 
\[\Rightarrow 0+mgR\left(1-{\cos 55{}^\circ \ }\right)=\frac{1}{2}mv^2+mgR\left(1-{\cos \theta \ }\right)\] 
\[\Rightarrow \frac{mv^2}{R}=2mg({\cos \theta \ }-{\cos 55{}^\circ \ })\] 
In part (a) we found that at any point the tension in the rope is 
\[T=mg{\cos \theta \ }+\frac{mv^2}{R}\] 
Where we have used the centripetal acceleration i.e. $a_r=v^2/R$. Combining the two relations give the tension of the rope as $T=mg(3\,{\cos \theta \ }-2\,{\cos 55{}^\circ \ })$. Set $T_{max}=980\ \mathrm{N}$ to find the desired angle $\theta \ $.
\[980=\left(65\times 9.8\right)\left(3\,{\cos \theta \ }-2\,{\cos 55{}^\circ \ }\right)\Rightarrow \theta =26.46{}^\circ \] 
 

(a) Since the system does not have any acceleration so is in a static situation. Calculate the torque acting on the system about the hinge point and set it zero.
\[\vec{\tau }=\vec{r}\times \vec{F}\Rightarrow \ \tau =mg\left(\frac{l}{2}\right){\cos 20{}^\circ \ }-Tl\,{\sin 20{}^\circ \ }=0\] 
\[\Rightarrow T=\frac{mg}{2}{\cot 20{}^\circ \ }=26.925\ \mathrm{kN}\] 
(b) The total force supplied by the hinge is the normal force that lies in the direction of the rod. if we decompose it relative t the $x$ and $y$ axes and apply the $\mathrm{\Sigma }\vec{F}=m\vec{a}$ to the center of mass of the rod (CM is motionless $s=0$), we get
\[\mathrm{\Sigma }\vec F_x=ma_x=0\Rightarrow N_x=T=26.925\ \mathrm{kN}\] 
\[\mathrm{\Sigma }\vec F_y=ma_y=0\Rightarrow N_y=mg=2000\times 9.8=19.6\ \mathrm{kN}\] 
\[\left|N\right|=\sqrt{N^2_x+N^2_y}=\sqrt{{\left(26925\right)}^2+{\left(19600\right)}^2}=33.303\ \mathrm{kN}\] 
 

(a) The two forces, the gravity and the normal force, act on the car as shown in the figure. The horizontal component of the normal force provides the centripetal force in the circular path. Apply Newton's 2${}^{nd}$ law to the car as follows
\[\mathrm{\Sigma }\vec F_x=mr{\omega }^2\Rightarrow N{\sin 30{}^\circ \ }=mr{\left(\frac{2\pi }{T}\right)}^2\] 
\[\mathrm{\Sigma }\vec F_y=ma_y=0\Rightarrow N{\cos 30{}^\circ \ }=mg\] 
Dividing these equations eliminates $N$ and allows us to find $r$: 
\[{\tan 30{}^\circ \ }=\frac{4{\pi }^2r}{T^2g}\Rightarrow r=\frac{gT^2{\tan 30{}^\circ \ }}{4{\pi }^2}=\frac{\left(9.8\right){\left(2\right)}^2}{4{\pi }^2}\,{\tan 30{}^\circ \ }=0.573\ \mathrm{m}\] 
(b) See part (a).
 

This is a static problem since there is no motion ($v=0\ ,\ a=0$). First draw a free body diagram then apply Newton's second law to the node in the figure as follows: \begin{gather*} \Sigma \vec F_x=0 \\ T_1-T_2{\cos \theta}=0 \\ \Rightarrow {\cos \theta}=\frac{T_1}{T_2} \end{gather*} Substituting the numerical value into this gives \begin{align*} \theta &={{\cos}^{-1} \left(\frac{T_1}{T_2}\right)} \\\\ &={{\cos}^{-1} \left(\frac{40}{50}\right)} \\\\ &=37^\circ \end{align*} Now write Newton's second law in the $y$ direction and find $M$ \begin{gather*} \Sigma \vec F_y=0 \\ T_2{\sin \theta}=Mg \end{gather*} Solving for $M$ and substituting the values into it yields \begin{align*} M&=\frac{T_2}{g}{\sin 37^\circ} \\\\ &=\frac{50}{9.8}(0.6) \\\\ &=3.06\,\rm kg \end{align*} In the above, we have used the well-known formula in trigonometry to find the ${\sin 37^\circ}$ \begin{gather*} {{\sin}^2 \theta}+{{\cos}^2 \theta}=1 \\\\ \Rightarrow {\sin \theta}=\sqrt{1-{{\cos}^2 \theta}} \\\\ =\sqrt{1-{(0.8)}^2}=0.6 \end{gather*} 

(a) Since $M$ is moving at a constant velocity so the acceleration of the system is zero.

Apply Newton's 2${}^{nd}$ law to each of the blocks as follows
\begin{gather*} \Sigma \vec F_y=0 \\ Mg-T_1=0 \\ \Rightarrow M=\frac{T_1}{g} \\\\ \Sigma \vec F_x=0 \\ T_1-f=0 \\ \Rightarrow T_1=\mu m_{12} g \end{gather*} 
Combining these two relations gives \[M=\mu m_{12}=0.3\times 12=3.6\,\rm kg\]
(b) In this case it is assumed the system has acceleration, so apply Newton's second law to the masses $M$ and $9\ \mathrm{kg}$ (see figures)
\begin{gather*} \Sigma \vec F_y=Ma \\ Mg-T_1=Ma \\ \Rightarrow T_1=M(g-a) \\\\ \Sigma \vec F_x=m_9 a \\ T_1-f=m_9 a \\ \Rightarrow T_1=m_9 a+\mu m_{12}g \end{gather*} Combining these two equations gives 
\begin{align*} M(g-a)&=m_9\,a+\mu m_{12} g \\\\ \Rightarrow a&=\frac{(M-\mu m_{12})}{m_9+M}g \\\\ &=\frac{5-0.3\times 12}{9+5}(9.8) \\\\ &=0.98\,\rm m/s^2 \end{align*} 

(a) Draw a free body diagram for the masses then apply the second law to them.

\begin{gather*} \Sigma \vec F_{1y}=0\to N=m_1g\,{\cos \theta \ } \\\\ \Sigma \vec F_{1x}=m_1a \\ \to T-m_1g\,{\sin \theta}-\underbrace{f_k}_{{\mu }_kN}=m_1a \\\\ \Sigma \vec F_{2y}=m_2a \to m_2g-T=m_2a\end{gather*}
In above, let the direction of the motion as the positive direction so we have supposed that the system moves in the clockwise. Combining these relations, we get the 
\begin{gather*} T= m_1\left(g{\sin \theta}+\mu_kg\,{\cos \theta}+a\right) \\ T=m_2(g-a) \end{gather*} Substitute the information given into the second equation; we will obtain the tension in the string as \[T=m_2(g-a)=3(9.8-2)=23.4\,\rm N\]
(b) This part is a question about coefficient of kinetic friction. Using the above equations, we can obtain the coefficient of kinetic friction $\mu_k$ \begin{gather*} m_1(g \sin \theta+\mu_k g\,\cos \theta+a)=m_2(g-a) \\\\ \mu_k=\frac{m_2 (g-a)-m_1 (g \sin\theta +a)}{m_1 g \cos\theta}=0.566\end{gather*}
(c) If we set $\mu_k=0$ in the curved bracket in the part (a), and solve for $a$, we will obtain \[a=\frac{(m_2-m_1\sin\theta)g}{m_1+m_1}\] By substituting the numerical values, we get \[a=\frac{3-2\sin 30^\circ}{3+2}\times 9.8=3.92\,\rm m/s^2\] 
 

(a) Apply Newton's second law to the blocks as

\begin{gather*} \Sigma \vec F_x=M_{tot}a \\ \to \, F{\cos 30^\circ}-f_2-f_1=(m_1+m_2)a \\\\ \Sigma\vec F_{2y}=0 \\ \to N_2+F\,{\sin 30^\circ }=m_2g \\\\ \Sigma \vec F_{1y}=0 \\ \to N_1=m_1g \end{gather*} We know that $f=\mu N$ gives the friction force so 
\[F{\cos 30^\circ}-\mu (m_2g-F{\sin 30^\circ})-\mu m_1g=(m_1+m_2)a\] 
Substituting the data given into the above expression, we obtain
\[\boxed{a=1.41\,\rm m/s^2}\]
Note: in the second law, all forces must be external forces. In the above, the tension on the string between the blocks is an internal force and does not contribute to calculating the total acceleration of the blocks.
(b) Now apply Newton's second law to one of the blocks, say $m_1$, as follows
\begin{gather*} \Sigma \vec F_{1x}=m_1a \\ \to T-\underbrace{\mu N_1}_{f_1}=m_1a \end{gather*}

Substituting the given date, we get
\begin{gather*} T=\mu m_1g+m_1a=m_1(\mu g+a) \\\\ \Rightarrow T=2(0.4\times 9.8+1.41)=10.66\,\rm N\end{gather*}