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Chapter 5: Problem 40

\(\bullet$$\bullet\) The coefficients of static and kinetic friction between a 476 \(\mathrm{N}\) crate and the warehouse floor are 0.615 and 0.420 , respectively. A worker gradually increases his horizontal push against this crate until it just begins to move and from then on maintains that same maximum push. What is the acceleration of the crate after it has begun to move? Start with a free-body diagram of the crate.

Short Answer

Expert verified
The acceleration after the crate begins moving is approximately 0.98 m/s虏.

Step by step solution

01

Analyze the forces on the crate

Begin by considering the forces acting on the crate. The forces include the gravitational force, normal force, the worker's pushing force, and frictional forces. The gravitational force can be calculated using the weight of the crate: 476 N. The normal force is equal to the gravitational force since there is no vertical acceleration.
02

Calculate static and kinetic frictional forces

The maximum static frictional force, which needs to be overcome for the crate to start moving, can be calculated using: \[ f_s = \mu_s \times N \] where \( \mu_s = 0.615 \) is the coefficient of static friction and \( N = 476 \text{ N} \) is the normal force. Calculate \( f_s = 0.615 \times 476 \). Then, calculate the kinetic frictional force using: \[ f_k = \mu_k \times N \] where \( \mu_k = 0.420 \) is the coefficient of kinetic friction.
03

Determine the net force after the crate starts moving

Once the crate begins moving, the worker's push equals the maximum static frictional force. After the movement begins, the force opposing the motion is the kinetic friction. Thus, the net force \( F_{\text{net}} \) is given by the difference between the pushing force (equals max static friction) and kinetic friction:\[ F_{\text{net}} = f_s - f_k \].
04

Calculate the acceleration of the crate

Use Newton's second law to determine the acceleration of the crate: \[ F_{\text{net}} = m \times a \] where \( m \) is the mass of the crate. First calculate \( m = \frac{W}{g} = \frac{476 \text{ N}}{9.8 \text{ m/s}^2} \). Substitute \( F_{\text{net}} \) from the previous step and solve for \( a \): \[ a = \frac{F_{\text{net}}}{m} \].

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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 prevents an object from moving when it is at rest. It needs to be overcome by an external force for motion to commence. This force acts opposite to the applied force on the object. In our scenario, the force of static friction is given by the equation
  • \( f_s = \mu_s \times N \)
where
  • \( \mu_s = 0.615 \) is the static friction coefficient,
  • and \( N = 476 \text{ N} \) represents the normal force exerted by the surface.
Thus, to calculate the maximum static friction force, which the worker must overcome, we multiply \( 0.615 \) by the weight of the crate, \( 476 \text{ N} \). This will give us the threshold force at which the crate begins to move. Static friction is generally larger than kinetic friction, helping prevent motion until a sufficient push is applied.
Kinetic Friction
Once the crate begins to move, static friction is no longer the opposing force. At this point, kinetic friction takes over, resisting the movement of the crate. Kinetic friction is generally less than static friction and can be calculated using the equation
  • \( f_k = \mu_k \times N \)
where
  • \( \mu_k = 0.420 \) is the kinetic friction coefficient,
  • and \( N = 476 \text{ N} \) is the normal force.
The kinetic frictional force will keep opposing the crate's movement as long as it is in motion. This creates a situation where the net force causing acceleration is reduced by the kinetic friction force, as the worker continues to apply the same force that overcame static friction.
Newton's Second Law
Newton's second law forms the foundation for understanding how the forces affect the crate鈥檚 acceleration. The law is given by the formula:
  • \( F = m \times a \)
where
  • \( F \) is the net force applied,
  • \( m \) is the mass of the object,
  • and \( a \) is the acceleration produced.
To find the acceleration, we first determine the mass of the crate using its weight and the acceleration due to gravity:
  • \( m = \frac{W}{g} = \frac{476 \text{ N}}{9.8 \text{ m/s}^2} \).
After overcoming static friction, the applied force switches to overcoming kinetic friction, calculating the net force:
  • \( F_{\text{net}} = f_s - f_k \).
From here, simply apply Newton's second law to solve for acceleration \( a \) by rearranging to \( a = \frac{F_{\text{net}}}{m} \). This process shows how effectively the force translates into motion once static friction is beaten.
Free-Body Diagram
A free-body diagram is a valuable tool used to visualize the forces acting on an object. For the crate, this diagram would include:
  • The gravitational force acting downward (weight of 476 N)
  • The normal force acting upward, equal in magnitude to the gravitational force
  • The applied force exerted by the worker horizontally
  • Frictional forces acting horizontally opposite to the applied force
Drawing a free-body diagram helps in organizing the forces and understanding their directions. This visualization assists in the transformation of vector forces into calculations needed to determine moments, especially the shift between static and kinetic friction. The diagram acts like a map of forces, simplifying complex equations into approachable and understandable elements for calculations.

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Most popular questions from this chapter

An "extreme" pogo stick utilizes a spring whose uncom- pressed length is 46 \(\mathrm{cm}\) and whose force constant is \(1.4 \times 10^{4} \mathrm{N} / \mathrm{m} . \mathrm{A} 60-\mathrm{kg}\) enthusiast is jumping on the pogo stick, compressing the spring to a length of only 5.0 \(\mathrm{cm}\) at the bottom of her jump. Calculate (a) the net upward force on her at the moment the spring reaches its greatest com- pression and (b) her upward acceleration, in \(\mathrm{m} / \mathrm{s}^{2}\) and \(g^{\prime}\) s at that moment.

\(\bullet\) You find that if you hang a 1.25 \(\mathrm{kg}\) weight from a vertical spring, it stretches 3.75 \(\mathrm{cm} .\) (a) What is the force constant of this spring in \(\mathrm{N} / \mathrm{m}\) ? (b) How much mass should you hang from the spring so it will stretch by 8.13 \(\mathrm{cm}\) from its original, unstretched length?

the task of designing an accelerometer to be used inside a rocket ship in outer space. Your equipment consists of a very light spring that is \(15.0 \mathrm{~cm}\) long when no forces act to stretch or compress it, \(\begin{array}{lllll}\text { plus } & \text { a } & 1.10 & \text { kg } & \text { weight. }\end{array}\) be attached to a friction- free tabletop, while the \(1.10 \mathrm{~kg}\) weight is attached to the other end, as shown in Figure 5.67 . (Such a spring-type accelerometer system was actually used in the ill-fated Genesis Mission, which collected particles of the solar wind. Unfortunately, because it was installed backward, it did not measure the acceleration correctly during the craft's descent to earth. As a result, the parachute failed to open and the capsule crashed on Sept. \(8,2004 .)\) (a) What should be the force constant of the spring so that it will stretch by \(1.10 \mathrm{~cm}\) when the rocket accelerates forward at \(2.50 \mathrm{~m} / \mathrm{s}^{2}\) ? Start with a free-body diagram of the weight. (b) What is the acceleration (magnitude and direction) of the rocket if the spring is compressed by \(2.30 \mathrm{~cm} ?\)

\(\bullet$$\bullet\) Atwood's Machine. A 15.0 -kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0 -kg counterweight is suspended from the other end of the rope, as shown in Fig. 5.60 . The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. (b) What is the magnitude of the upward acceleration of the load of bricks? (c) What is the ten- sion in the rope while the load is moving? How does the ten- sion compare to the weight of the load of bricks? To the weight of the counterweight?

\(\bullet$$\bullet\) Prevention of hip injuries. People (especially the elderly) who are prone to falling can wear hip pads to cushion the impact on their hip from a fall. Experiments have shown that if the speed at impact can be reduced to 1.3 \(\mathrm{m} / \mathrm{s}\) or less, the hip will usually not fracture. Let us investigate the worst-case sce- nario, in which a 55 kg person completely loses her footing (such as on icy pavement) and falls a distance of \(1.0 \mathrm{m},\) the dis- tance from her hip to the ground. We shall assume that the per- son's entire body has the same acceleration, which, in reality, would not quite be true. (a) With what speed does her hip reach the ground? (b) A typical hip pad can reduce the person's speed to 1.3 \(\mathrm{m} / \mathrm{s}\) over a distance of 2.0 \(\mathrm{cm} .\) Find the acceleration (assumed to be constant) of this person's hip while she is slow- ing down and the force the pad exerts on it. (c) The force in part (b) is very large. To see if it is likely to cause injury, calcu- late how long it lasts.
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