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Physics Bootcamp

Samuel J. Ling

Section 6.20 Forces Bootcamp

Exercises Exercises

Second Law

1. Finding Mass Ratios From Acceleration Ratios and Inventing Units of Mass.
Follow the link: Example 6.20.

Third Law

2. Forces on the Car and the Tow Truck.
Follow the link: Example 6.23.

Net Force

3. Net Force on a Box Pulled by Two Forces in Perpendicular Directions - Example.
Follow the link: Example 6.27.
5. Net Force of Two Forces General Case.
Follow the link: Example 6.28.
6. Net Force of Three Forces Simple Case.
Follow the link: Example 6.29.
7. Three Balanced Forces - Find Magnitude and Direction of One.
Follow the link: Exercise 6.6.2.2.
8. Four Balanced Forces - Find Magnitudes of Two.
Follow the link: Exercise 6.6.2.3.

Weight

9. Weight of a Dumb-bell-Shaped Object.
Follow the link: Example 6.32.

Normal Force

10. A Box on Floor Pushed Against a Wall.
Follow the link: Example 6.35.
11. Normal Force on a Book Against the Wall.
Follow the link: Exercise 6.10.1.

Static Friction Force

12. Determining Static Friction.
Follow the link: Example 6.43.

Kinetic Friction Force

15. Kinetic Friction - Accelerating.
Follow the link: Example 6.52.
17. Motion on an Inclined Plane with Kinetic Friction.
Follow the link: Exercise 6.12.2.

Drag Force

18. Computing Viscous Drag Force Given the Speed.
Follow the link: Example 6.63.
19. Computing Inertial Drag Force.
Follow the link: Example 6.64.
20. Drag Force and the Terminal Velocity.
Follow the link: Example 6.65.
21. Comparison of Inertial Drag Forces on Two Airplanes.
Follow the link: Exercise 6.14.5.1.
22. (Calculus) Speed and Distance on a Skier with Inertial Drag Force.
Follow the link: Exercise 6.14.5.2.
23. (Calculus) Practice with a Friend: Accounting for Air Drag on a Bob Sled.
Follow the link: Exercise 6.14.5.3.

Spring Force

24. A Block Hanging from a Spring - The Stretch of Spring.
Follow the link: Example 6.70.

Coupled Motion

28. Force Between Two Books Moving Together.
Follow the link: Example 6.85
29. Force Between Two Books Stacked with One Book Pushed from Side.
Follow the link: Exercise 6.17.3.1
30. A Block Pushed on Another Block and Held in Place by Static Friction.
Follow the link: Exercise 6.17.3.2
31. Two Blocks Moving Vertically Coupled Through an Ideal Pulley.
Follow the link: Example 6.91
32. Two Blocks Coupled Through an Ideal Pulley, one Block Moving Horizontally.
Follow the link: Exercise 6.17.3.3
33. Practice with a Friend: Coupled Motion of Two Blocks on Two Sides of a Fixed Wedge.
Follow the link: Exercise 6.17.3.7
35. Practice with a Friend: Block on Incline of an Accelerating Wedge.
Follow the link: Exercise 6.17.3.8
37. Block not Sliding in Motion of Three Blocks and a Pulley.
Follow the link: Exercise 6.17.3.6

Dynamics of Circular Motion

40. Car Rounding a Turn on a Slippery Banked Road.
Follow the link: Example 6.115
43. Practice with a Friend: How Hard Does the Toy Car Press on the Track?
Follow the link: Exercise 6.18.3
44. A Person at the Wall of a Spinning Drum.
Follow the link: Exercise 6.18.1
45. Circular Motion of a Block Supported by a Hanging Mass.
Follow the link: Exercise 6.18.4

Dynamics in Polar Coordinates

47. (Calculus) Rock Tied to String Moving in a Circle.
Follow the link: Example 6.128
48. (Calculus) Equation of Motion of a Simple Pendulum.
Follow the link: Exercise 6.19.1
49. (Calculus) Movement of a Bead on a Spinning Frictionless Rod.
Follow the link: Example 6.130
50. (Calculus) A Block Attached to a String Rotated on a Table with Length of String Changing.
Follow the link: Exercise 6.19.2
51. Practice with a Friend: (Calculus) Length of String Changing at Constant Acceleration.
Follow the link: Exercise 6.19.4
52. Practice with a Friend: (Calculus) Angular Speed and Tension of a Simple Pendulum.
Follow the link: Exercise 6.19.3

Miscellaneous

53. Balancing Spring Force by Static Friction.
A \(30\text{-kg}\) box on a smooth floor is attached to a spring of spring constant \(100\text{ N/m}\) and pulled horizontally by an increasing force \(F \) until the point when the box starts to slide. At that instant, \(F = 75\text{ N}\text{.}\) The coefficient of static friction between the bottom of the box and the floor surface is \(0.2\text{.}\)
At the instant when box is about to slide, by how much is the spring stretched?
Figure 6.137. Figure for Exercise 6.20.53
Hint.
Balance forces acting on the block.
Answer.
\(0.161\text{ m}. \)
Solution.
We start by identifying forces on the block as in Figure 6.138. Note that static friction has its maximum value at the instant of interest.
Figure 6.138. Figure for Exercise 6.20.53 (solution).
Since acceleration of the block is zero, forces on the block are balanced. Since every force is along one or the other axis, it is very easy to write equations of motion along the two axes directly. Lets also just use \(F_{s}^{\text{max}} = \mu_s F_N \text{.}\) Also, lets use \(F_{sp} = k \Delta l\text{.}\) We need to find \(\Delta l\text{.}\)
\begin{align*} \amp F - \mu_s F_N - k\Delta l = 0, \\ \amp F_N - mg = 0, \end{align*}
Solve the second equation for \(F_N \) and use that in the first equation, which we solve for \(\Delta l\text{.}\)
\begin{equation*} \Delta l = \dfrac{1}{k} \left( F -\mu_smg \right). \end{equation*}
Putting the numerical values in this equation we get
\begin{equation*} \Delta l = \dfrac{1}{100} \left( 75 -0.2\times 30 \times 9.81 \right) = 0.161\text{ m}. \end{equation*}
54. An Athlete Pulling on A Rope - Balancing Forces to Find Minimum \(\mu_s\).
A \(100\text{-kg}\) athlete pulls a rope attached to a strong pole at an angle of \(15^{\circ}\) from horizontal such that the tension in the rope devlops to \(800\text{ N}\text{.}\)
Figure 6.139.
(a) Find nomal and frictional forces on the athlete by the floor.
(b) What must be minimum value of the coefficient of static friction between the sole of the shoes of the athlete and the floor so that the athlete does not slip on the floor while he is pulling on the rope?
Hint.
Draw a Free-Body Diagram of the athlete.
Answer.
(a) Friction = \(773\text{ N}\text{,}\) Normal = \(774\text{ N}\text{.}\) (b) \(1.0 \text{..}\)
Solution 1. a
We start by identifying forces on the block in Figure 6.140, where I have used symbols for the forces: \(W \) for weight, \(T \) for weight, \(N_L \) for the normal on the left foot, \(N_R \) for the normal on the right foot, \(S_L \) for the static friction on the left foot, and \(S_R \) for the static friction on the right foot.
Figure 6.140. Figure for Exercise 6.20.54(a).
In the translational motion, we can combine the normal forces into one and call that \(N=N_L+N_R\text{,}\) and similary for the static friction, \(S=S_L+S_R\text{.}\) We will have to keep them separate when we look at rotation.
Since acceleration of the block is zero, the forces are balanced. Since only \(T \) is not along one of the axes, we will need to work out its components. Let \(\theta \) stand for angle \(15^{\circ}\text{.}\)
\begin{equation*} T_x = T\,\cos\,\theta,\ \ T_y = T\,\sin\,\theta. \end{equation*}
Now, we can write the equations of motion along the two axes.
\begin{align*} \amp T\,\cos\,\theta - S = 0, \\ \amp T\,\sin\,\theta + N - W = 0, \end{align*}
Here, the values of \(T \text{,}\) \(\theta\text{,}\) and \(W=mg\text{,}\) are known. Therefore, we solve the first equation to get \(S \text{,}\) and the second one for \(N \text{.}\)
\begin{align*} \amp S = T\,\cos\,\theta = 800 \times \cos\,15^{\circ} = 773\text{ N}, \\ \amp N = W - T\,\sin\,\theta = 100\times 9.81 - 800 \times \sin\,15^{\circ} = 774\text{ N}. \end{align*}
Solution 2. b
The definition of \(\mu_s \) requires the maximum static friction.
\begin{equation*} \mu_s = \dfrac{S^{\text{max}}}{N}. \end{equation*}
From the problem description, we need the minimum \(\mu_s \text{.}\) That means, if we assume the \(S \) we found was the \(S^{\text{max}}\text{,}\) then the \(\mu_s \) we will get will be the minimum required for the static condition to hold.
\begin{equation*} \mu_s^{\text{min}} = \dfrac{S^{\text{found}}}{N} = \dfrac{773}{774} = 1.0. \end{equation*}
55. Motion of One Pulley and Two Blocks on One Pulley.
Figure 6.141 shows a two pulley two block system. While pulley \(\text{P}_1\) only rotates with its center remaining fixed, \(\text{P}_2\) rotates as well as moves. Assume both pulleys to be ideal (massless and frictionless) so that tension on the two sides of each pulley have the same magnitude. Do not assume tensions in the two strings are equal but you can assume that the length of strings do not change during the motion. Find the tensions in the two strings and accelerations of the two blocks.
Figure 6.141.
Hint.
Use an upward pointed \(y\) axis and write lengths of the two strings in terms of the \(y\) coordinates. From there deduce relation among accelerations of the two blocks.
Answer.
Partial: \(a_{1y} = \left( \frac{2m_2 - m_1}{m_1 + 4 m_2} \right)\; g \) if positive \(y\) axis pointed up.
Solution.
Let us use upward pointed \(y\) axis as in Figure 6.142 to analyze the motion. Let \(y_1\text{,}\) \(y_p\text{,}\) \(y_2\) denote the positions of the centers of block 1, pulley \(P_2\text{,}\) and block 2 respectively. We will need \(y_p\) to figure out the relation between changes in \(y_1\) and \(y_2\text{.}\)
Figure 6.142.
Let us look at each string to find relations among changes in these \(y\)’s. To facilitate this, let \(y_\text{top}\) be the \(y\) of the fixed pulley. Let us also denote lengths of the strings by \(l_1\) and \(l_2\) respectively. Then, we have
\begin{align*} \amp \left( y_\text{top} - y_1 \right) + ( y_\text{top} - y_p ) = l_1 = \text{constant}\\ \amp \left( y_p \right) + ( y_p - y_2 ) = l_2 = \text{constant} \end{align*}
Therefore, we have the following relations among the changes in the coordinates.
\begin{align*} \amp \Delta y_1 + \Delta y_p = 0\\ \amp 2 \Delta y_p - y_2 = 0 \end{align*}
From this, we get
\begin{equation} v_{2y} = -2v_{1y},\ \ a_{2y} = -2a_{1y}.\tag{6.78} \end{equation}
This relation between the accelerations and \(y\) equations of motion of the two blocks and the moving pulley (zero mass) will be sufficient to answer the questions about accelerations of the two blocks and the tensions. Using the force diagrams of each block we have the following equations.
\begin{align*} \amp T_1 - m_1 g = m_1 a_{1y}\\ \amp T_2 - m_2 g = m_2 a_{2y}\\ \amp T_1 - 2T_2 = 0 \end{align*}
Using Eq. (6.78), we find
\begin{align*} \amp a_{1y} = \left( \frac{2m_2 - m_1}{m_1 + 4 m_2} \right)\; g,\ \ \ a_{2y} = -2a_{1y}. \\ \amp T_1 = \left( \frac{6m_1m_2}{m_1 + 4 m_2} \right)\; g,\ \ \ T_2 = \frac{1}{2}T_1. \end{align*}
56. Practice with a Friend: Motion of One Pulley and Three Blocks on One Pulley.
Figure 6.143 shows a two pulley three block system. While pulley \(\text{P}_1\) only rotates with its center remaining fixed, \(\text{P}_2\) rotates as well as moves. Assume both pulleys to be ideal (massless and frictionless) so that tension on the two sides of each pulley have the same magnitude. Do not assume tensions in the two strings are equal but you can assume that the length of strings do not change during the motion. Find the tensions in the two strings and accelerations of the three blocks.
Figure 6.143.
Answer Hint: when \(m_1=m_2=m_3\text{,}\) \(|a_1| = \frac{1}{3}g\text{.}\) Try this case first before you solve the general problem.
57. Movement of Three Blocks on Three Pulleys one of Which is Also Moving.
Three masses are connected to two fixed pulleys and a moving pulley as shown in the figure. Assume all pulleys massless and frictionless and all strings massless. Find the accelerations of the three masses.
Figure 6.144.
Solution.
To work out the constraint in the motion of the coupled system of masses and pulleys, let us work with a coordinate system with the origin at the level of the line joining A and C and the direction of the positive \(y\)-axis pointed down as shown in Figure 6.145. Let the \(y\)-coordinates of \(m_1\text{,}\) \(m_2\text{,}\) \(M\text{,}\) and B be \(y_1\text{,}\) \(y_2\text{,}\) \(y_3\text{,}\) and \(y_p\) respectively.
Figure 6.145.
There are two constraints, namely the length of the string over the pulleys is constant and the distance between M and B is fixed. These physical constraints becomes
\begin{align*} \amp y_1 +\pi R + y_p+\pi R +y_p + \pi R + y_3 = \textrm{constant}.\\ \amp y_3 - y_p = \textrm{constant}. \end{align*}
By taking successive derivatives with respect to time, we get the following constraints among the \(y\)-components of the velocities and accelerations.
\begin{align} \amp v_{1y} +2v_{2y} + v_{3y} = 0.\tag{6.79}\\ \amp a_{1y} +2a_{2y} + a_{3y} = 0.\tag{6.80} \end{align}
With the positive \(y\)-axis pointed down, we obtain the following \(y\)-components of the equations of motion for \(m_1\text{,}\) \(m_2\text{,}\) and the system constaining the mass \(M\) and the center pulley.
\begin{align} \amp m_1 g - T = m_1 a_{1y}.\tag{6.81}\\ \amp m_2 g - T = m_2 a_{2y}.\tag{6.82}\\ \amp Mg - 2T = M a_{3y}.\tag{6.83} \end{align}
To make use of Eq. (6.80), divide each equation here by masses.
\begin{align} \amp g - \frac{1}{m_1} T = a_{1y}.\tag{6.84}\\ \amp g - \frac{1}{m_2} T = a_{2y}.\tag{6.85}\\ \amp g - \frac{2}{M} T = a_{3y}.\tag{6.86} \end{align}
Multiply the last equation by 2 and add the three equations allows us to use Eq. (6.80) to arrive at
\begin{equation} 4g-\left[ \frac{1}{m_1} + \frac{2}{m_2} +\frac{2}{M} \right] T = 0.\tag{6.87} \end{equation}
which gives the following for the tension in the string.
\begin{equation} T = 4 \left[ \frac{1}{m_1} + \frac{2}{m_2} +\frac{2}{M} \right] ^{-1}\ g.\tag{6.88} \end{equation}
We can put this \(T\) in Eqs. (6.84) - (6.86) to obtain the \(y\)-component of the accelerations.
58. Rope Swinging in a Conical Path at Constant Speed.
A rope of mass \(M\) and length \(L\) is swung uniformly with the angular speed \(\omega\) in a circle that makes an angle \(\theta\) with the vertical as shown in Figure 6.146 . What is the tension in the rope at a distance \(b\) from the top? Note: each element of the rope moves in a horizontal circle with uniform circular motion.
Figure 6.146.
Answer.
\(\frac{M \omega^2}{2L} \left(L^2 - b^2 \right).\)
Solution.
Consider a small element of the rope between \(r\) and \(r+\Delta r\) from the suspension point. This element of mass \(\Delta m = (M/L)\Delta r\) will be moving in a circle of radius \(R = r \sin\theta\) at the speed \(v = \omega r\text{.}\)
The free-body diagram of the forces on the element shows that there are three forces on it: the weight \(\Delta m g\) vertically down, the tension at \(r\) pulling the element towards the suspension point and the tension at \(r+\Delta r\) pulling the element away from the suspension. The acceleration of the element is pointed horizontally towards the center of the circle of motion.
Figure 6.147.
Let us denote the tension at \(r\) by \(T(r)\text{.}\) The horizontal component of the equation of motion is then
\begin{equation*} T(r)\sin\theta - T(r+\Delta r) \sin\theta = \Delta m \frac{v^2}{R}. \end{equation*}
Writing the right side in \(r\) gives
\begin{equation*} T(r)\sin\theta - T(r+\Delta r) \sin\theta = \left( \frac{M}{L}\right)\Delta r \omega^2\sin\theta\ r. \end{equation*}
Dividing both sides by \(\Delta r\) and taking the limit \(\Delta r \rightarrow 0\) limit transforms the left side into a derivative:
\begin{equation*} \frac{dT}{dr} = \lim_{\Delta r\rightarrow 0}{\frac{T(r+\Delta r) - T(r)}{\Delta r}}, \end{equation*}
and we arrive at the following rule for the rate of the change of the tension along the rope.
\begin{equation*} \frac{dT}{dr} = - \frac{M \omega^2}{L} \ r. \end{equation*}
To find the tension we need to integrate this equation. We have seen this type of equation when the velocity was a linearly increasing function of time and we needed the displacement. Similar technique can be used here. We notice that \(r=L\) corresponds to the free end of the rope at the end and the tension there will be zero. Therefore, let us integrate from \(r=L\) to \(r=r\) corresponding to \(T=0\) to \(T=T(r)\text{.}\)
\begin{equation*} \int_0^{T(r)} dT = - \frac{M \omega^2}{L} \int_L^r\ r\ dr. \end{equation*}
This gives
\begin{equation*} T(r) = \frac{M \omega^2}{2L} \left(L^2 - r^2 \right). \end{equation*}
For \(r=b\) this is
\begin{equation*} T_b = \frac{M \omega^2}{2L} \left(L^2 - b^2 \right). \end{equation*}
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