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Chapter 5: Problem 88

Block \(B\) has mass \(5.00 \mathrm{~kg}\) and sits at rest on a horizontal, frictionless surface. Block \(A\) has mass \(2.00 \mathrm{~kg}\) and sits at rest on top of block \(B\). The coefficient of static friction between the two blocks is \(0.400 .\) A horizontal force \(\vec{P}\) is then applied to block \(A .\) What is the largest value \(P\) can have and the blocks move together with equal accelerations?

Short Answer

Expert verified
The maximum force \(P\) that can be applied to block A without causing slippage between the blocks can be calculated using the formula \(P = 渭_s * m_a * g / (1 + 渭_s)\).

Step by step solution

01

Identifying the Forces

Identify the forces acting on each block: weight (due to gravity), normal force, applied force \(P\), and frictional force.
02

Calculate Total Force

The total force (F_total) on block A will be the sum of the applied force (\(P\)) and the frictional force. From Newton's 2nd Law, F_total = mass * acceleration. So \(P + f_{s,max} = m_a * a\) (1) where \(m_a\) is the mass of block A, \(a\) is the acceleration, and \(f_{s,max}\) is the maximum static friction.
03

Calculate the maximum static friction

The maximum static friction \(f_{s,max}\) between the two blocks is given by \(f_{s,max} = 渭_s * N\), where \(渭_s\) is the coefficient of static friction and N is the normal force. The normal force is equal to the weight of block A, which is \(m_a * g\). So we can rewrite the equation as \(f_{s,max} = 渭_s * m_a * g\) (2), where \(g\) is the acceleration due to gravity.
04

Calculate Frictional force acting on Block B

In addition to the above, for block B, there's no external force applied, so the net force is the frictional force alone. \(F_total = f_{s,max}\). So, \(m_b * a = f_{s,max}\) (3), where \(m_b\) is the mass of block B.
05

Solving for P

From equation (1), (2) and (3), it's possible to solve for \(P \). We equate \(f_{s,max}\) from equation (2) and equation (3), and then substitute in equation (1) to solve for \(P\). The solution to these equations gives \(P = 渭_s * m_a * g - m_b * a\). However, since both blocks move with the same acceleration, we can solve the system of equations substituting \(m_b * a\) by \(P\) in the above equation. This provides us the equation \(P = 渭_s * m_a * g - P\). Solving for \(P\) gives \(P = 渭_s * m_a * g / (1 + 渭_s)\)

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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 it is in contact with another surface. In this problem, static friction acts between blocks A and B. It's crucial because it prevents block A from sliding off block B when the force \( \vec{P} \) is applied to block A. The static frictional force can be calculated using the formula:
  • Maximum static friction \( f_{s,\text{max}} = \mu_s \times N \)
The coefficient of static friction \( \mu_s \) is given as 0.400, and it determines how much force is needed before block A starts sliding. The normal force \( N \) is the force exerted by block B on A, perpendicular to the surface. It's equal to the weight of block A, which is its mass \( m_a = 2.00 \text{ kg} \) times the gravitational acceleration \( g \) (approximately \( 9.81 \text{ m/s}^2 \)). By calculating \( f_{s,\text{max}} \), we understand the threshold that the applied force \( \vec{P} \) must not exceed for both blocks to move together smoothly without slipping.
Forces on Objects
Understanding how forces act on objects is essential in solving this problem. Block A experiences gravity, a normal force from block B, and the horizontal applied force \( \vec{P} \). Meanwhile, block B, resting on a frictionless surface, experiences a normal force from the ground and the force due to the weight of block A. The forces are balanced vertically because of these normal forces.
In a horizontal plane, however, static friction becomes significant. If \( \vec{P} \) overcomes the static friction, block A will slide over block B. Newton's Second Law states that the total force is equal to the mass of an object times its acceleration (\( F = m \times a \)).
  • For block A: \( P = m_a \times a + f_{s,\text{max}} \)
  • For block B: The net force is actually \( f_{s,\text{max}} \) since it helps both blocks accelerate together.
By understanding and equating these forces, we can derive the condition required for both blocks to maintain the same acceleration.
Acceleration
Acceleration refers to the rate of change of velocity of an object. When calculating the forces on blocks A and B, their acceleration is crucial as it determines the motion. In this context, acceleration can be found by solving the equation derived from equating the forces.
Since both blocks move together, they have equal acceleration. Using Newton's second law, the formula becomes:
  • \( m_b \times a = f_{s,\text{max}} \)
This allows us to find the effective acceleration of block B due to the static friction. The same friction ensures block A doesn't slide, as long as its calculation holds true. Acceleration here provides insight into the dynamics of the motion where both blocks mimic a single system, all due to the force \( \vec{P} \) applied. The system's acceleration can be maximized by finding suitable conditions in the interaction of the given forces and employing friction effectively.

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Most popular questions from this chapter

Friction in an Elevator. You are riding in an elevator on the way to the 18 th floor of your dormitory. The elevator is accelerating upward with \(a=1.90 \mathrm{~m} / \mathrm{s}^{2} .\) Beside you is the box containing your new computer; the box and its contents have a total mass of \(36.0 \mathrm{~kg} .\) While the elevator is accelerating upward, you push horizontally on the box to slide it at constant speed toward the elevator door. If the coefficient of kinetic friction between the box and the elevator floor is \(\mu_{\mathrm{k}}=0.32,\) what magnitude of force must you apply?

Forces During Chin-ups. When you do a chin-up, you raise your chin just over a bar (the chinning bar), supporting yourself with only your arms. Typically, the body below the arms is raised by about \(30 \mathrm{~cm}\) in a time of \(1.0 \mathrm{~s},\) starting from rest. Assume that the entire body of a \(680 \mathrm{~N}\) person doing chin-ups is raised by \(30 \mathrm{~cm},\) and that half the \(1.0 \mathrm{~s}\) is spent accelerating upward and the other half accelerating downward, uniformly in both cases. Draw a free-body diagram of the person's body, and use it to find the force his arms must exert on him during the accelerating part of the chin-up.

When a crate with mass \(25.0 \mathrm{~kg}\) is placed on a ramp that is inclined at an angle \(\alpha\) below the horizontal, it slides down the ramp with an acceleration of \(4.9 \mathrm{~m} / \mathrm{s}^{2} .\) The ramp is not friction less. To increase the acceleration of the crate, a downward vertical force \(\overrightarrow{\boldsymbol{F}}\) is applied to the top of the crate. What must \(F\) be in order to increase the acceleration of the crate so that it is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? How does the value of \(F\) that you calculate compare to the weight of the crate?

A physics major is working to pay her college tuition by performing in a traveling carnival. She rides a motorcycle inside a hollow, transparent plastic sphere. After gaining sufficient speed, she travels in a vertical circle with radius \(13.0 \mathrm{~m}\). She has mass \(70.0 \mathrm{~kg},\) and her motorcycle has mass \(40.0 \mathrm{~kg} .\) (a) What minimum speed must she have at the top of the circle for the motorcycle tires to remain in contact with the sphere? (b) At the bottom of the circle, her speed is twice the value calculated in part (a). What is the magnitude of the normal force exerted on the motorcycle by the sphere at this point?

A \(2.00 \mathrm{~kg}\) box is suspended from the end of a light vertical rope. A time-dependent force is applied to the upper end of the rope, and the box moves upward with a velocity magnitude that varies in time according to \(v(t)=\left(2.00 \mathrm{~m} / \mathrm{s}^{2}\right) t+\left(0.600 \mathrm{~m} / \mathrm{s}^{3}\right) t^{2}\). What is the tension in the rope when the velocity of the box is \(9.00 \mathrm{~m} / \mathrm{s} ?\)
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