Problem 29 - University Physics with Modern Physics

Question

A 45.0 -kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and observe that the crate just begins to move when your force exceeds 313 \(\mathrm{N}\) . After that you must reduce your push to 208 \(\mathrm{N}\) to keep it moving at a steady 25.0 \(\mathrm{cm} / \mathrm{s}\) (a) What are the coefficients of static and kinetic friction between the crate and the floor? (b) What push must you exert to give it an acceleration of 1.10 \(\mathrm{m} / \mathrm{s}^{2} ?\) (c) Suppose you were performing the same experiment on this crate but were doing it on the moon instead, where the acceleration due to gravity is 1.62 \(\mathrm{m} / \mathrm{s}^{2}\) . (i) What magnitude push would cause it to move? (ii) What would its acceleration be if you maintained the push in part (b)?

Step-by-Step Solution

Answer

(a) \( \mu_s \approx 0.71 \), \( \mu_k \approx 0.47 \); (b) 257.5 N; (c)(i) 51.76 N, (ii) 4.96 m/s².

Step 1: Identify Given Information

We have a crate with mass \( m = 45.0 \text{ kg} \). The force required to initiate movement is \( F_s = 313 \text{ N} \) and the force required to maintain constant velocity is \( F_k = 208 \text{ N} \). We will find the coefficients of static friction \( \mu_s \) and kinetic friction \( \mu_k \).

Step 2: Calculate the Normal Force

The normal force \( N \) on an object resting on a horizontal surface is equal to its weight. Thus, \( N = mg \), where \( g = 9.8 \text{ m/s}^2 \). Therefore, \( N = 45.0 \times 9.8 = 441 \text{ N} \).

Step 3: Find Coefficient of Static Friction

The coefficient of static friction \( \mu_s \) can be found using the formula \( F_s = \mu_s N \). Solving for \( \mu_s \), we get: \[ \mu_s = \frac{F_s}{N} = \frac{313}{441} \approx 0.71 \]

Step 4: Find Coefficient of Kinetic Friction

The coefficient of kinetic friction \( \mu_k \) can be found using the formula \( F_k = \mu_k N \). Solving for \( \mu_k \), we get: \[ \mu_k = \frac{F_k}{N} = \frac{208}{441} \approx 0.47 \]

Step 5: Calculate Force for a Given Acceleration

To find the force \( F \) required to accelerate the crate at \( a = 1.10 \text{ m/s}^2 \), use the equation: \[ F = ma + F_k = 45.0 \times 1.10 + 208 \] \[ F = 49.5 + 208 = 257.5 \text{ N} \]

Step 6: Moon Experiment - Calculate Moon Normal Force and Static Friction

On the moon, the gravitational acceleration is \( g = 1.62 \text{ m/s}^2 \). The normal force becomes \( N = mg = 45 \times 1.62 = 72.9 \text{ N} \). Thus, \( F_s = \mu_s N = 0.71 \times 72.9 \approx 51.76 \text{ N} \).

Step 7: Moon Experiment - Calculate Acceleration with Given Force

Using the force \( F = 257.5 \text{ N} \) (from part b) on the moon and the kinetic friction, we find \( F_k = \mu_k N = 0.47 \times 72.9 \approx 34.26 \text{ N} \). The net force \( F_{net} = F - F_k = 257.5 - 34.26 = 223.24 \text{ N} \).\[ a = \frac{F_{net}}{m} = \frac{223.24}{45} \approx 4.96 \text{ m/s}^2 \]

Key Concepts

Static Friction

Static friction is the resistance that keeps an object at rest when a force is applied to it. It must be overcome to set the object in motion. This frictional force opposes the initial movement of the object, maintaining it stationary until the applied force is sufficiently high.
In the exercise provided, static friction is what kept the 45.0-kg crate from moving until we applied a force greater than 313 N.
The formula to calculate static friction is \[ F_s = \mu_s N \]where:

• \( F_s \) is the force of static friction,
• \( \mu_s \) is the coefficient of static friction,
• \( N \) is the normal force.

By using the known maximum static force and the normal force of 441 N, we found the static friction coefficient to be approximately 0.71 in this example.

Kinetic Friction

Once an object is in motion, kinetic friction comes into play. This is the force opposing the object's movement as it slides over a surface. Unlike static friction, kinetic friction remains constant once the object starts moving. It is usually less than static friction for the same pair of surfaces.
For the 45.0-kg crate, after it started moving, the kinetic friction required to maintain its motion was met with a 208 N force.
The equation for kinetic friction is:\[ F_k = \mu_k N \]where:

• \( F_k \) is the force of kinetic friction,
• \( \mu_k \) is the coefficient of kinetic friction,
• \( N \) is the normal force.

Through calculation, the coefficient for kinetic friction in this case turned out to be approximately 0.47.

Normal Force

The normal force is an essential concept in understanding friction. It acts perpendicular to the surface on which an object rests. This force balances the object's weight when on a flat surface, thereby preventing it from falling through. In our scenario, the crate's normal force is equivalent to its weight, calculated as the product of mass and gravitational acceleration.
For the crate with a mass of 45.0 kg on Earth:\[ N = mg \]\[ N = 45.0 \times 9.8 = 441 \text{ N} \] This formula becomes crucial when calculating both static and kinetic frictions, because these frictions depend directly on the normal force.

Acceleration

Acceleration is the rate at which the velocity of an object changes. It reflects how quickly an object speeds up, slows down, or changes direction. In the exercise, we needed to apply additional force to accelerate the crate at a rate of 1.10 m/s².
The force required to achieve this acceleration, after considering kinetic friction, is calculated using Newton's second law:\[ F = ma + F_k \]where:

• \( F \) is the total force applied,
• \( m \) is the mass of the object (45 kg),
• \( a \) is the desired acceleration (1.10 m/s²),
• \( F_k \) is the kinetic friction (208 N).

Hence, the total initial force required to accelerate the crate is 257.5 N.

Gravitational Acceleration

Gravitational acceleration is the acceleration due to the force of gravity. On Earth, this acceleration is approximately 9.8 m/s². However, it varies depending on the celestial body.
In the moon experiment, gravitational acceleration changed to 1.62 m/s², altering the normal force and subsequently the forces of static and kinetic frictions.
This lower gravitational pull meant a smaller force was sufficient to move the crate on the moon compared to Earth.

• Normal force on moon: \( N = 45 \times 1.62 = 72.9 \text{ N} \)
• Maximum static force on moon: \( F_s = \mu_s N \approx 51.76 \text{ N} \)

Understanding gravitational acceleration is crucial as it directly impacts how forces like friction depend on weight, varying as gravity varies.