Problem 28 - University Physics with Modern Physics

Question

A box of bananas weighing 40.0 \(\mathrm{N}\) rests on a horizontal surface. The coefficient of static friction between the box and the surface is \(0.40,\) and the coefficient of kinetic friction is 0.20 . (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 \(\mathrm{N}\) to the box and the box is initially at rest? (c) What minimum horizontal force must the monkey apply to start the box in motion? (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? If the monkey applies a horizontal force of \(18.0 \mathrm{N},\) what is the magnitude of the friction force and what is the box's acceleration?

Step-by-Step Solution

Answer

(a) 0 N; (b) 6.0 N; (c) 16.0 N; (d) 8.0 N; (18 N applied force) friction force: 8.0 N, acceleration: 2.45 m/s².

Step 1: Calculating Static Friction with No External Force

The static friction force when the box is at rest and no external force is applied is equal to the gravitational force acting on it. Thus, the frictional force is equal to the normal force. Since the box weighs 40.0 N, the frictional force exerted on the box is 0 N because it is at rest and no external force is trying to move it.

Step 2: Calculating Friction Force with 6 N Applied

When a horizontal force of 6.0 N is applied, we check if the box remains at rest or moves. The maximum static friction force is \( f_s = \mu_s \times N = 0.40 \times 40.0 = 16.0 \) N. Since 6.0 N is less than 16.0 N, the static friction force equals the applied force, i.e., 6.0 N.

Step 3: Finding Minimum Force to Initiate Motion

To set the box into motion, the monkey needs to overcome the maximum static friction force. Therefore, the minimum force is equal to the maximum static friction force, which is 16.0 N.

Step 4: Force Needed to Maintain Constant Velocity

Once the box is in motion, kinetic friction takes over. The force needed to keep the box moving at constant velocity is equal to the kinetic friction force. This is given by \( f_k = \mu_k \times N = 0.20 \times 40.0 N = 8.0 \) N.

Step 5: Analyzing 18 N Applied Horizontal Force

With an 18.0 N force applied, the box overcomes static friction and moves. The friction force is kinetic, calculated as 8.0 N. The acceleration can be found using Newton's second law. The net force is \( 18.0 - 8.0 = 10.0 \) N. The box's mass is \( m = \frac{40.0}{9.8} \approx 4.08 \) kg, and the acceleration \( a = \frac{10.0}{4.08} \approx 2.45 \,m/s^2 \).

Key Concepts

Static Friction

Static friction is the type of friction that keeps an object at rest when a force is applied. It acts in the opposite direction to the force trying to move the object. This frictional force exists until the applied force becomes strong enough to overcome it. For instance, in the case of a box weighing 40 N, which is at rest on a surface, the static friction works to prevent movement until a threshold force is achieved. If you apply a force that's less than this threshold, the box remains in place.

The static frictional force can be calculated using the formula:

• \( f_s = \mu_s \times N \)

where \( \mu_s \) is the coefficient of static friction, and \( N \) is the normal force. For example, with a coefficient of static friction of 0.40 and a normal force of 40 N, the static friction is:

• \( f_s = 0.40 \times 40.0 = 16.0 \) N

If a force less than this maximum frictional force is applied (e.g., 6.0 N), the frictional force equals the applied force, and the box remains at rest.

Kinetic Friction

Once an object starts moving, it is subjected to kinetic friction, which tends to slow it down. Kinetic friction operates slightly differently from static friction in that it typically requires less force to maintain movement compared to what was needed to start the motion. In the context of our box, once it begins moving, kinetic friction takes over from static friction.

The formula to calculate kinetic friction is:

• \( f_k = \mu_k \times N \)

where \( \mu_k \) is the coefficient of kinetic friction. For the box example with a coefficient of kinetic friction of 0.20:

• \( f_k = 0.20 \times 40.0 = 8.0 \) N

This means a force of 8.0 N is required to maintain the box moving at a constant velocity. So, if once the box is moving, the monkey has to apply this 8 N of force to keep it sliding steadily.

Newton's Second Law

Newton's second law provides a fundamental explanation of how forces affect the motion of objects. It states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. This can be expressed mathematically as:

• \( F = m \times a \)

where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is the acceleration.

In our box example, once a force of 18.0 N is applied and motion starts, the net force acting on the box is calculated by subtracting the kinetic friction from the applied force:

• Net Force \( = 18.0 \- 8.0 = 10.0 \) N

To find the box's acceleration, use the mass of the box \( m \approx 4.08 \) kg:

• \( a = \frac{10.0}{4.08} \approx 2.45 \text{ m/s}^2 \)

This shows how the box accelerates at around 2.45 m/s² due to the force applied by the monkey.

Forces

Forces are influences that cause changes in the motion of objects. They are vector quantities, meaning they possess both magnitude and direction. There are numerous types of forces including gravitational, frictional, and applied forces. In our context with the box:

• *Gravitational Force:* It acts downward, equivalent to the weight of the box (40 N).
• *Normal Force:* This force acts perpendicular to the surface, balancing the gravitational force when the box is at rest.
• *Frictional Forces:* These include static and kinetic friction, acting opposite to the direction of movement or applied force.
• *Applied Force:* Such as the 6.0 N or 18.0 N force exerted by the monkey to move the box.

Understanding how these forces interact helps determine the conditions necessary for the movement of objects, such as not only triggering an initial shift in the box but also maintaining motion.