Problem 40 - College Physics

Question

\(\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.

Step-by-Step Solution

Answer

The acceleration after the crate begins moving is approximately 0.98 m/s².

Step 1: 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.

Step 2: 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.

Step 3: 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:\[ F_{\text{net}} = f_s - f_k \].

Step 4: 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 \( a \): \[ a = \frac{F_{\text{net}}}{m} \].

Key Concepts

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’s 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.