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Chapter 4: Problem 83

Two blocks are stacked on a frictionless table, and a horizontal force \(F\) is applied to the top block (block 1). Their masses are \(m_{1}=2.50 \mathrm{~kg}\) and \(m_{2}=3.75 \mathrm{~kg}\). The coefficients of static and kinetic friction between the blocks are 0.456 and 0.380 , respectively a) What is the maximum applied force \(F\) for which \(m_{1}\) will not slide off \(m_{2} ?\) b) What are the accelerations of \(m_{1}\) and \(m_{2}\) when \(F=24.5 \mathrm{~N}\) is applied to \(m_{1}\) ?

Short Answer

Expert verified
To find the maximum force, use the formula: $$ F_{max} = f_{s,max} = 渭_s m_{1}g. $$ Plug in the values for \(渭_s\), \(m_{1}\), and \(g\). b) What are the accelerations of the two blocks when a force of 24.5 N is applied to the top block? To find the accelerations of the two blocks, first find the kinetic friction force: $$ f_k = 渭_k m_{1}g. $$ Then, find the net forces acting on each block: $$ F_{net,1} = F - f_k \quad \text{and} \quad F_{net,2} = f_k. $$ Finally, use Newton's second law to find the accelerations: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Calculate the values for \(a_{1}\) and \(a_{2}\).

Step by step solution

01

1. Calculate the maximum static friction force between the blocks

The maximum static friction force \(f_{s,max}\) between the two blocks is given by the product of the coefficient of static friction \(渭_s\) and the normal force \(N\). Since there is no vertical acceleration, the normal force between the blocks must equal the weight of block 1: \(N = m_{1}g\). Thus, $$ f_{s,max} = 渭_s m_{1}g, $$ where \(g = 9.81 \,\text m/s^2\) is the gravitational acceleration.
02

2. Find the maximum force F

For block 1 to not slide off block 2, the applied force should not exceed the maximum friction force between the blocks. Therefore, $$ F_{max} = f_{s,max} = 渭_s m_{1}g. $$ Plug in the values for \(渭_s\), \(m_{1}\), and \(g\) to find the maximum applied force \(F_{max}\).
03

3. Calculate the kinetic friction force between the blocks

When a force of \(F=24.5\,\text N\) is applied to block 1, the kinetic friction force \(f_k\) between the blocks needs to be calculated. Since the force is less than the maximum force calculated in step 2, the blocks will move together. The kinetic friction force can be calculated by the product of the coefficient of kinetic friction \(渭_k\) and the normal force: $$ f_k = 渭_k N = 渭_k m_{1}g. $$
04

4. Calculate the total force acting on each block

The total net force acting on block 1, \(F_{net,1}\), will be the applied force minus the kinetic friction force: $$ F_{net,1} = F - f_k. $$ The total net force acting on block 2, \(F_{net,2}\), will be equal to the kinetic friction force: $$ F_{net,2} = f_k. $$
05

5. Calculate the accelerations of the two blocks

Using Newton's second law, the acceleration of each block can be found as follows: $$ a_{1} = \frac{F_{net,1}}{m_{1}} \quad \text{and} \quad a_{2} = \frac{F_{net,2}}{m_{2}}. $$ Plug in the values for \(F_{net,1}\), \(F_{net,2}\), \(m_{1}\), and \(m_{2}\) to find the accelerations \(a_{1}\) and \(a_{2}\) of the blocks.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Static Friction
Static friction is the force that keeps an object at rest when a force is applied, preventing it from moving. When two surfaces are in contact, static friction acts parallel to the surface of contact. It's a balancing act鈥攊t resists the initial motion, making it crucial to understanding how objects begin to move.
In the given exercise, the static friction is what keeps block 1 from sliding off block 2 even when a horizontal force is applied.
  • The maximum static friction force (\(f_{s, max}\)) is calculated by multiplying the coefficient of static friction (\(渭_s\)) by the normal force (\(N\)).
  • In this scenario, since there鈥檚 no vertical movement, the normal force is equal to the gravitational force on block 1: \(N = m_{1}g\).
  • The equation to find the force is \(f_{s, max} = 渭_s m_{1}g\).
Understanding static friction helps us recognize that as long as the applied force stays below this maximum frictional force, block 1 stays put on block 2.
Kinetic Friction
Kinetic friction comes into play when an object is already moving over a surface. It鈥檚 usually less than static friction, meaning it takes less force to keep an object moving than to start it moving. While static friction prevents motion, kinetic friction acts in the opposite direction of the movement, working to slow it down.
In the exercise, once block 1 overcomes static friction with the applied force, kinetic friction keeps influencing the motion between block 1 and block 2.
  • Kinetic friction force (\(f_k\)) can be calculated by the formula: \(f_k = 渭_k m_{1}g\).
  • The coefficient of kinetic friction (\(渭_k\)) determines how much friction exists while the object is in motion.
  • In the situation given, the applied force is less than the maximum static friction force, so both blocks will move together, with kinetic friction working between them.
Realizing this distinction between static and kinetic friction is crucial in solving problems involving sliding objects.
Acceleration
Acceleration is the rate at which an object's velocity changes over time. Newton's Second Law of Motion tells us that acceleration happens when a net force is applied to an object, expressed in the equation \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration.
In this particular problem, once the force overcomes static friction and is less than maximum static friction, both blocks accelerate together. Determining their acceleration involves:
  • Calculating the net force acting on each block.
  • For block 1, the net force (\(F_{net,1}\)) is the applied force minus the kinetic friction: \(F_{net,1} = F - f_k\).
  • For block 2, the net force (\(F_{net,2}\)) is simply the kinetic friction force since it is propelled by the friction from block 1: \(F_{net,2} = f_k\).
  • Using \(F = ma\), solve for acceleration: \(a_1 = \frac{F_{net,1}}{m_1}\) and \(a_2 = \frac{F_{net,2}}{m_2}\).
Grasping how acceleration works in this perfect interplay of forces and mass is vital to determine how systems of objects behave when subjected to forces.

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Most popular questions from this chapter

A bowling ball of mass \(M_{1}=6.0 \mathrm{~kg}\) is initially at rest on the sloped side of a wedge of mass \(M_{2}=9.0 \mathrm{~kg}\) that is on a frictionless horizontal floor. The side of the wedge is sloped at an angle of \(\theta=36.9^{\circ}\) above the horizontal. a) With what magnitude of horizontal force should the wedge be pushed to keep the bowling ball at a constant height on the slope? b) What is the magnitude of the acceleration of the wedge, if no external force is applied?

A mass slides on a ramp that is at an angle of \(\theta\) above the horizontal. The coefficient of friction between the mass and the ramp is \(\mu\). a) Find an expression for the magnitude and direction of the acceleration of the mass as it slides up the ramp. b) Repeat part (a) to find an expression for the magnitude and direction of the acceleration of the mass as it slides down the ramp.

-4.45 A pinata of mass \(M=8.0 \mathrm{~kg}\) is attached to a rope of negligible mass that is strung between the tops of two vertical poles. The horizontal distance between the poles is \(D=2.0 \mathrm{~m},\) and the top of the right pole is a vertical distance \(h=0.50 \mathrm{~m}\) higher than the top of the left pole. The pinata is attached to the rope at a horizontal position halfway between the two poles and at a vertical distance \(s=1.0 \mathrm{~m}\) below the top of the left pole. Find the tension in each part of the rope due to the weight of the pinata.

What coefficient of friction is required to stop a hockey puck sliding at \(12.5 \mathrm{~m} / \mathrm{s}\) initially over a distance of \(60.5 \mathrm{~m} ?\)

4.25 You have just joined an exclusive health club, located on the top floor of a skyscraper. You reach the facility by using an express elevator. The elevator has a precision scale installed so that members can weigh themselves before and after their workouts. A member steps into the elevator and gets on the scale before the elevator doors close. The scale shows a weight of 183.7 lb. Then the elevator accelerates upward with an acceleration of \(2.43 \mathrm{~m} / \mathrm{s}^{2},\) while the member is still standing on the scale. What is the weight shown by the scale's display while the elevator is accelerating?
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