Inclined Planes: Practice Problems with Solutions
In this article, we explore the fascinating world of inclined planes in physics with a number of solved practice problems designed for high school and college students. Get ready to sharpen your problem-solving skills and deepen your knowledge of this fundamental concept in physics!
Inclined Plane Problems without Friction:
Problem (1): A 45 kg object rests on a 30° inclined plane.
(a) What force is exerted down the inclined plane?
(b) What force acts perpendicular to the plane?
Solution: In all inclined plane problems, the first step is to choose a suitable coordinate system. It is common to choose the positive direction along and down the plane so that the up direction is perpendicular to the plane.
Two forces always act on an object resting on a smooth inclined plane (without friction): weight force and a force perpendicular to the slope, which is called the normal force.
The normal force always acts along the up direction of our chosen coordinate system (tilted coordinate system), but the weight force should be decomposed into that adopted coordinate on the inclined plane.
As you can see in the figure below, one of its components is parallel to the plane downward, and the other is perpendicular and in the $-y$ direction.
(a) With these brief explanations, the downward weight component parallel to the slope always pulls an object down on an inclined plane. \begin{align*} F_x&=mg\sin\theta \\ &=45\times 9.8\sin 30^\circ \\ &=\boxed{220.5\,\rm N} \end{align*}
(b) As you noticed along the perpendicular direction to the plane, two forces act on the object while it has no motion (i.e., $a_y=0$). Applying Newton's second law to this direction gives us the normal force on the object \begin{gather*} F_{net-y}=ma_y \\\\ F_N-w_{\bot}=0 \\\\ F_N-mg\cos\theta=0 \\\\ \Rightarrow F_N=\underbrace{mg\cos\theta}_{w_{\bot}} \end{gather*} Substituting the given numerical values into this, we have \[\boxed{F_N=45\times 9.8 \cos 30^\circ=382\,\rm N}\]
Problem (2): A car of mass $m$ is at the top of an icy 15° slope whose length is 30 m.
(a) Find the acceleration of the car?
(b) How long does it take for the car to reach the bottom?
(c) How fast is the car at the bottom?
Solution: The slope is icy, so it has no friction. Recall from the previous problem that the downward weight component parallel to the inclines is $mg\sin\theta$. This force is the cause of the car going down from the top of the inclined plane.
(a) Because the incline is frictionless, the only force that acts on the car is $w_{\parallel}=mg\sin\theta$. Applying Newton's second law and substituting the numerical value yields \begin{gather*} F_{net}=ma_x \\\\ mg\sin\theta=ma_x \\\\ \rightarrow a_x=g\sin\theta \\\\ a_x=(9.8) \sin 15^\circ \\\\ \Rightarrow \boxed{a_x=2.6\,\rm m/s^2} \end{gather*} As this equation shows, the acceleration along the inclined plane does not depend on the object's mass.
(b) The car starts its motion from rest, i.e., $v_0=0$. The car's acceleration is also found in the previous part. The only kinematic equation that relates these known values together is $\Delta x=\frac 12 at^2+v_0t$, where $\Delta x$ is the total displacement traveled by the object along the slope, which here is $\Delta x=30\,\rm m$. \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 30=\frac 12 (2.6)t^2+0 \\\\ \rightarrow t=\sqrt{\frac{2\times 30}{2.6}} \\\\ \Rightarrow \boxed{t=4.8\,\rm s} \end{gather*}
(c) To find the velocity at the bottom, we can use either of two kinematic equations $v^2-v_0^2=2a\Delta x$ or $v=v_0+at$. We choose the first equation and put the numbers into it \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-0=2(2.6)(30) \\\\ \Rightarrow \boxed{v=12.48\,\rm m/s} \end{gather*}
Problem (3): A hockey puck with an acceleration of $6\,\rm m/s^2$ is sliding down a $12-\rm m$-long frictionless ramp.
(a) What is the angle of the ramp?
(b) How long does it take the puck to reach the bottom?
(c) If the mass of the puck is doubled, how does its acceleration change down the incline?
Solution: The ramp or inclined plane has an unknown angle and must be determined. Assume the positive $x$-axis to be down the ramp.
(a) The only force that pulls the puck down the slope and accelerates it is the component of weight force parallel to the inclined plane, $w_{\parallel}=mg\sin\theta$. Therefore, using the second law, we have \begin{gather*} F_{net}=ma_x \\\\ mg\sin\theta=ma_x \\\\ \Rightarrow a_x=g\sin\theta \end{gather*} Substituting the numerical values into this, solving for angle $\theta$, and taking the inverse sine of both sides, will get \begin{align*} \theta &=\sin^{-1} \left(\frac{a_x}{g}\right) \\\\ &=\sin^{-1} \left(\frac{6}{9.8}\right) \\\\ &=37.7^\circ \end{align*}
(b) The puck is initially held at rest, $v_0$, and the ramp's length $\Delta x=12\,\rm m$ is also known, so use the following kinematic equation to find the required time. \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 12=\frac 12 (6)t^2+0 \\\\ \Rightarrow \boxed{t=2\,\rm s} \end{gather*}
(c) In part (a), you saw that the acceleration of an object sliding down a frictionless ramp depends on $g$ and the ramp's angle $\theta$, not the mass $m$. Therefore, by doubling the puck's mass, its acceleration does not change along the inclined plane. \[a_{new}=a_{old}=6\,\rm m/s^2\]
Problem (4): A $1500-\rm kg$ car is at rest on a frictionless hill inclined at $17^\circ$.
(a) Sketch a free-body diagram and calculate the components $w_{\parallel}$, $w_{\bot}$, and the normal force acting on the car.
(b) Find the car's acceleration when it is released.
(c) If the hill is $55-\rm m$-long, what would be the car's final velocity at the bottom of the hill?
(d) How long will it take for the car to reach the end of the hill?
Solution: (a) Two forces act on the car. The normal force perpendicular to the hill, $F_N$, and the weight force, $w=mg$. The weight force has two components that are parallel $w_{\parallel}$ and perpendicular $w_{\bot}$ to the inclined plane.
From the following free-body diagram, we can calculate these components as \begin{align*} w_{\parallel} &=mg\sin\theta \\ &=(1500\times 9.8) \sin 17^\circ \\ &=\boxed{4297.86\,\rm N} \\\\ w_{\bot}&=mg\cos\theta \\&=1500\times 9.8\cos 17^\circ \\&=\boxed{14057.68\,\rm N} \end{align*}
(b) The car perpendicular to the hill does not have acceleration, $a_y=0$, but parallel to the hill is accelerating $a_x=?$ which must be determined.
Applying Newton's second law along the $x$- and $y$-axes on the hill, get \begin{gather*} F_{net-x}=ma_x \\ mg\sin\theta=ma_x \\ \Rightarrow a_x=g\sin\theta \\\\ F_{net-y}=ma_y \\ F_N-w_{\bot}=0 \\\Rightarrow F_N=w_{\bot}=\boxed{14057.68\,\rm N} \end{gather*} Therefore, the car's acceleration is \[a_x=9.8\times \sin 17^\circ =2.86\,\rm m/s^2 \]
(c) This part is related to a kinematics problem in physics. The car is displaced along the hill by $\Delta x=55\,\rm m$ from rest, $v_0=0$. Its acceleration is also found in the previous part, $a_x=2.86\,\rm m/s^2$. Using the following kinematics equation and solving for $v$, gets \begin{gather*} v^2-v_0^2=2a\Delta x \\\\ v^2-0=2(2.86)(55) \\\\ \Rightarrow \boxed{v=17.73\,\rm m/s} \end{gather*}
(d) Apply the equation $v=v_0+at$ and solve for the unknown time $t$ as below \begin{gather*} v=v_0+at \\ 17.73=0+(2.86)t \\ \Rightarrow \boxed{6.2\,\rm s} \end{gather*}
Problem (5): A box with a mass of $45\,\rm kg$ is held at rest by a rope on a frictionless ramp inclined at $30^\circ$. Draw a free-body diagram and find the value of the tension force exerted by the rope.
Solution: The box is at rest, so there is no motion or acceleration along the $x$- and $y$- axes parallel and perpendicular to the inclined plane, respectively.
As the free-body diagram shows, the tension force is upward on the ramp and the weight component $mg\sin\theta$, as always, down the incline. Applying Newton's second law gives us \begin{gather*} F_{net-x}=ma_x \\\\ T-mg\sin\theta=0 \\\\ \Rightarrow T=mg\sin\theta \end{gather*} By substituting the given values into this, the value of tension force is obtained as follows \[T=45\times 9.8\times \sin 30^\circ=\boxed{220.5\,\rm N}\]
Inclined Plane Problems with Friction:
Problem (6): A slippery object slides down a ramp of $9\,\rm m$ long inclined at $10^\circ$. The kinetic friction between the object and the ramp is $\mu_k=0.06$. How long does it take for the object to reach the bottom?
Solution: The only force exerted on the object and forcing it to go down is the component of weight parallel to the inclined plane. The kinetic friction force up the incline opposes the weight force down the incline. Balancing these two exerting forces gives acceleration to the slippery object.
Consider the downward direction of the ramp as the positive $x$ direction. Applying Newton's second law to forces along the inclined plane yields \begin{gather*} F_{net-x}=ma_x\\\\ mg\sin\theta-f_k=ma_x \\\\ mg\sin\theta -\mu_k N=mg \end{gather*} where in the last line, we used the definition of friction force. Applying the above law along the direction normal to the slope, given that in this direction the object does not move, we can find the normal force on the object \begin{gather*} F_N-mg\cos\theta=0 \\ \Rightarrow F_N=mg\cos\theta \end{gather*} Substituting this into the first expression and removing the common factor $m$, we get the object's acceleration along the rough inclined plane \begin{align*} a_x &=g(\sin\theta-\mu_k \cos\theta) \\\\ &=(9.8) \left(\sin 10^\circ-(00.6) \cos 10^\circ \right) \\\\ &=1.12\,\rm m/s^2 \end{align*} Assume the object was initially at rest, $v_0=0$. On the other hand, the total displacement or distance that the object covered is the length of the incline, $\Delta x=9\,\rm m$.
By knowing that information, we can use the displacement kinematic equation $\Delta x=\frac 12 at^2+v_0t$ to find the required time as below \begin{gather*} \Delta x=\frac 12 at^2+v_0t \\\\ 9=\frac 12 (1.12)t^2 + 0 \\\\ \rightarrow t=\sqrt{\frac{2\times 9}{1.12}} \\\\ \Rightarrow \quad \boxed{t=4\,\rm s} \end{gather*} Thus, our slippery object takes about $4$ seconds to reach the bottom of the inclined plane.
Problem (7): A $54.6\,\rm kg$ block is resting on a $7^\circ$ slope.
(a) What is the normal force acting on the block?
(b) What is the friction force holding the block on the slope?
Solution: except for the weight and friction forces, no other external forces at angles, act on the block resting on the inclined plane.
(a) Recall that, as always in all inclined plane problems, in the presence of these two forces, the normal force on an object on a slope is $F_N=mg\cos\theta$. Substituting the given values gives us \[F_N=54.6\times 9.8\times \cos 7^\circ=531\,\rm N\]
(b) The block is at rest on the slope, so there should be a force to oppose the weight component $mg\sin\theta$ to hold the block motionless on the inclined plane. This opposing force is the same friction force.
Consequently, in this special case, the friction force is found to be $f=mg\sin\theta$ whose magnitude is also as below \[f=54.6\times 9.8\times \sin 7^\circ= 65.2\,\rm N\]
Problem (8): A $46.5\,\rm kg$ crate is resting motionless on a ramp inclined at an angle of $20^\circ$. What is the coefficient of friction between the crate and the ramp?
Solution: The crate does not have acceleration along the ramp, so the downward weight component $mg\sin\theta$ must be balanced with the upward friction force on the incline. Therefore, we have $f=mg\sin\theta$. On the other hand, in these situations, the normal force acting on the object also has a magnitude of $F_N=mg\cos\theta$.
From the problems on the coefficient of static friction force, we remember that $f=\mu F_N$. Combining everything together, we get \begin{gather*} f=\mu F_N \\\\ mg\sin\theta=\mu (mg\cos \theta) \\\\ \rightarrow \mu=\frac{\sin\theta}{\cos\theta}=\tan\theta \end{gather*} Substituting the numerical values into this equation, we get \[\boxed{\mu=\tan 20^\circ=0.36} \]
This question has an important note for us. It tells us that when an object is either at rest (motionless) or moving with a constant velocity along a rough inclined plane, the coefficient of friction between the incline and the object is independent of the object's mass and only depends on the angle of the incline.
Problem (9): A box weighing $95\,\rm N$ is held at rest on a slope inclined at an angle of $25^\circ$.
(a) Find the magnitudes of the weight forces parallel and perpendicular to the inclined plane?
(b) Assuming the coefficient of static friction between the box and the incline is $0.55$, what is the maximum frictional force?
(c) Does the box move? If yes, at what rate does the box move down? If not, with what $\mu_s$ will the box be on the verge of sliding?
Solution: The box's weight and the angle of incline $\theta=25^\circ$ are given.
(a) As you learned up to now, the weight force components parallel ($w_\parallel$) and perpendicular ($w_\bot$) to the incline are given as below \begin{align*} w_\parallel&=mg\sin\theta \\&=95\times \sin 25^\circ \\ &=\boxed{40.14\,\rm N} \\\\ w_\bot&=mg\cos\theta \\&=95\cos 25^\circ \\&=\boxed{86\,\rm N} \end{align*}
(b) Static friction is the maximum frictional force that opposes motion and holds an object from moving. It is also defined as $f_{s,max}=\mu_s F_N$, where $F_N$ is the normal force exerted on an object sitting on the incline.
Recall that in the case of an external force, applying to the object at an angle, the magnitude of the normal force is calculated as below: \begin{align*} f_{s,max}&=\mu_s (mg\cos\theta) \\ &=(0.55)(95) \cos 25^\circ \\ &=\boxed{47.35\,\rm N} \end{align*}
(c) To realize whether the box moves down the incline or remains at rest, we must compare the weight component $w_\parallel$ down the incline with the maximum static friction upward along the incline.
In the last two parts, we find that \[\underbrace{w_\parallel}_{40.14} < \underbrace{f_{s,max}}_{47.35} \] The force opposing the motion is greater than the force pulling down the box along the inclined plane, consequently, the box remains motionless.
For the box to start sliding along from the top of the incline, the parallel weight component $w_\parallel$ must be equal to the frictional force. Equating these two forces and solving for $\mu_s$, yield the required coefficient of static friction to put the box on the verge of sliding. \begin{gather*} w_\parallel =f_{s,max} \\ mg\sin\theta =\mu_s (mg\cos\theta) \\\\ \rightarrow \mu_s=\tan\theta \\ \Rightarrow \boxed{\mu_s=\tan 25^\circ=0.466}\end{gather*}
Problem (10): A box of mass $35\,\rm kg$ sits on a $30^\circ$ incline. Consider these two following cases and find out what is being said.
(a) If the box does not move down the incline, what can you say about the static frictional force acting on the box?
(b) If the box slides down the incline with an acceleration of $3\,\rm m/s^2$, find the kinetic frictional force and the coefficient of kinetic friction between the box and the incline.
Solution: Like any other inclined plane practice problem, first find the parallel and perpendicular components of the object's weight along the incline as below \begin{align*} w_\parallel&=mg\sin\theta \\&=35\times 9.8 \times \sin 30^\circ \\ &=\boxed{171.5\,\rm N} \\\\ w_\bot&=mg\cos\theta \\&=35 \times 9.8 \cos 30^\circ \\&=\boxed{297.0\,\rm N} \end{align*} Now we want to examine the behavior of the box along the slope with/without the frictional force.
(a) If the box is not going to slide down the incline, $w_\parallel$ must be equal to the static frictional force $f_s$. \[f_s = w_\parallel \Rightarrow \boxed{f_s = 171.5\,\rm N}\] This is the actual static friction force required to keep the box at rest on the incline.
(b) Now the conditions are changed, and the box accelerates down the incline. The frictional force applied during the movement is called the kinetic friction force, $f_k=\mu_k F_N$, where $F_N=mg\cos\theta$. The box is moving down the incline, so the friction must be upward on the incline.
Choose the downward direction of the incline as the positive $x$-axis and perpendicular to the slope to be the positive $y$-axis.
Applying Newton's second law and solving for $f_k$, we will have \begin{gather*} f_{net}=ma_x \\ mg\sin\theta-f_k=ma_x \\ 35\times 9.8\times \sin 30^\circ-f_k=(35)(3) \\\\ \Rightarrow \boxed{f_k=66.5\,\rm N} \end{gather*} The kinetic friction is defined as the normal force times the coefficient of kinetic force, $f_k=\mu_k F_N$. Solving this for $\mu_k$, get \[\mu_k=\frac{f_k}{F_N}=\frac{66.5}{297.0}=\boxed{0.22}\]
Problem (11): A $5-\rm kg$ box is placed on a surface with a kinetic coefficient of friction of $0.28$. The surface is slowly raised and reaches an angle of $48^\circ$. What would be the acceleration of the box down the incline?
Solution: In the earlier questions, we saw that the acceleration of an object on a rough inclined plane is written as \[a=g(\sin\theta-\mu_k \cos\theta)\] Substituting the numerical values into it gives \begin{align*} a&=(9.8) \left(\sin 48^\circ-(0.28)\cos 48^\circ \right) \\&=\boxed{5.44\,\rm m/s^2} \end{align*}
Problem (12): A car weighing $2\times 10^4\,\rm N$ is at rest at the bottom of a $22^\circ$ hill. Several people are pushing the car uphill with a combined force of $8500\,\rm N$ parallel to the incline. (Assume the hill has negligible friction.)
(a) What acceleration does the car obtain?
(b) The people push on the car for $15\,\rm s$. How far does the car go up the incline?
Solution: In the following free-body diagram, all forces acting on the car along with their corresponding components are shown. As always the weight components are found as below \begin{align*} w_{\parallel}&=mg\sin\theta \\&= (2\times 10^4) \sin 22^\circ \\ &=\boxed{7492.13\,\rm N} \\\\ w_{\bot}&=mg\cos\theta \\ &=(2\times 10^4) \cos 22^\circ \\&=\boxed{18543.67\,\rm N} \end{align*} Take the direction of the acceleration as the positive $x$-direction. In this case, the weight component $w_{\parallel}$ is in the negative $x$ direction or downhill, but the external force, exerted by people, is uphill.
(a) Applying the second law along the $x$-direction (on the inclined plane) gives us \begin{gather*} F_{net-x}=ma_x \\ F_{ext}-w_{\parallel}=ma_x \\ 8500-7492.13=2040.81 a_x \\ \Rightarrow \boxed{a_x=0.49\,\rm m/s^2} \end{gather*}
(b) The car is initially at rest at the bottom of the hill. The only kinematic relationship that connects our known values, $a=0.49\,\rm m/s^2$ and $t=15\,\rm m$, is $\Delta x=\frac 12 at^2+v_0t$, where $\Delta x$ is the distance traveled on the incline. \begin{align*} \Delta x&=\frac 12 at^2+v_0t \\\\ &=\frac 12 (0.49)(15)^2+0 \\\\ &=\boxed{55.125\,\rm m} \end{align*}
Problem (13): A box weighing $175\,\rm N$ rests on a horizontal surface with a static coefficient of friction of $\mu_s=0.85$.
(a) At what angle should the surface be raised so that the box would just start to move?
(b) Assume the kinetic coefficient of friction is $\mu_k=0.42$. At what angle should the ramp be lowered so that the box moves without acceleration?
Solution: On the horizontal surface, the only forces initially acting on the box are the normal $F_N$ and the weight $w=mg$ forces. As soon as it makes an angle with the level, no longer the weight pulls the box perpendicularly.
At this point, we must choose a coordinate system on the inclined plane, as you saw in the previous problems.
(a) Prior to reaching a specific angle, the box is at rest yet. This is because the static friction force upward the incline is greater than $mg\sin\theta$ downward along the incline. As the angle increases, the weight component parallel to the incline $mg\sin\theta$ gradually increases in magnitude until it equals the maximum static friction force $f_{s,max}$. At this moment, the box is on the verge of slipping downward. \begin{gather*} f_{s,max}=mg\sin\theta \\ \mu_s F_N=mg\sin\theta \\ \mu_s (mg\cos\theta)=mg\sin\theta \\ \Rightarrow \boxed{\mu_s=\tan \theta} \end{gather*} Thus, the desired angle is arctangent (inverse of tangent) of the static coefficient of friction $\mu_s$ \begin{align*} \theta&=\arctan{\mu_s} \\ &=\arctan{0.85} \\ &=\boxed{40.36^\circ} \end{align*}
(b) As you have seen in part (a), the box is at rest until the surface reaches an angle of about $40^\circ$. Beyond that angle, the box would accelerate downslope. Now, by reducing the angle we can make a situation in which the box would travel with constant velocity (no acceleration). This occurs when the weight force component $mg\sin\theta$ down slope becomes equal to the kinetic friction force $f_k=\mu_k F_N$ upslope. Equating these two forces will give us the desired angle \begin{gather*} f_k=mg\sin\theta \\ \mu_k (mg\cos\theta)=mg\sin\theta \\ \Rightarrow \boxed{\tan\theta=\mu_k} \end{gather*} Substituting the given value into this equation and taking the inverse tangent of both sides, we get \begin{gather*} \tan\theta=0.42 \\ \rightarrow \theta =\tan^{-1}(0.42)=\boxed{22.7^\circ} \end{gather*}
Summary:
In this long article, we solved some problems on inclined planes with and without frictional force. In both cases, we learned that the normal force acting perpendicularly on an object on an inclined plane is written as $F_N=mg\cos\theta$.
On the other hand, the weight force component $w_{\parallel}$ parallel to the incline pulls the object downslope whose magnitude is $w_{\parallel}=mg\sin\theta$.
Author: Dr. Ali Nemati
Page Published: Sep 18, 2022
