Problem 49 - College Physics

Question

\(\bullet$$\bullet\) A crate of 45.0 -kg tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it and ob- serve that the crate just begins to move when your force ex- ceeds 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) Static friction coefficient: 0.709, kinetic friction coefficient: 0.471. (b) Push for acceleration: 257.5 N. (c) On the Moon: (i) Push: 51.7 N, (ii) Acceleration: 4.96 m/s².

Step 1: Calculate Normal Force

To find the frictional forces, we first need to find the normal force. On Earth, the force of gravity on the crate is given by \[ F_g = m imes g = 45.0 \text{ kg} imes 9.81 \text{ m/s}^2 = 441.45 \text{ N} \]Since the crate is resting on a horizontal floor, the normal force \( F_N \) is equal to the gravitational force, so \( F_N = 441.45 \text{ N} \).

Step 2: Determine Static Friction Coefficient

The static friction force \( f_s \) is the force needed to start moving the crate, which is given as 313 N. The formula for static friction is:\[ f_s = \mu_s \times F_N \]Solve for \( \mu_s \):\[ \mu_s = \frac{f_s}{F_N} = \frac{313 \text{ N}}{441.45 \text{ N}} \approx 0.709 \]

Step 3: Determine Kinetic Friction Coefficient

The kinetic friction force \( f_k \) keeps the crate moving at a constant speed and is given as 208 N. The formula for kinetic friction is:\[ f_k = \mu_k \times F_N \]Solve for \( \mu_k \):\[ \mu_k = \frac{f_k}{F_N} = \frac{208 \text{ N}}{441.45 \text{ N}} \approx 0.471 \]

Step 4: Calculate Push for Acceleration

To find the force needed to accelerate the crate at 1.10 \( \text{m/s}^2 \), use Newton's second law:\[ F = m \cdot a + f_k \]Where \( a \) is the acceleration and \( f_k \) is the kinetic friction force. Substitute to find \( F \):\[ F = 45.0 \text{ kg} \times 1.10 \text{ m/s}^2 + 208 \text{ N} = 49.5 \text{ N} + 208 \text{ N} = 257.5 \text{ N} \]

Step 5: Recalculate Normal Force on the Moon

On the Moon, the gravity is \( 1.62 \text{ m/s}^2 \). Calculate the new normal force:\[ F_{N,\text{moon}} = m \times g_{\text{moon}} = 45.0 \text{ kg} \times 1.62 \text{ m/s}^2 = 72.9 \text{ N} \]

Step 6: Magnitude of Push to Move the Crate on the Moon

Using the static friction coefficient calculated earlier:\[ f_{s,\text{moon}} = \mu_s \times F_{N,\text{moon}} = 0.709 \times 72.9 \text{ N} = 51.7 \text{ N} \]

Step 7: Acceleration on the Moon with Push from Part B

Using the kinetic friction coefficient:\[ f_{k,\text{moon}} = \mu_k \times F_{N,\text{moon}} = 0.471 \times 72.9 \text{ N} = 34.3 \text{ N} \]Use net force to calculate acceleration:\[ F - f_{k,\text{moon}} = m \times a \Rightarrow 257.5 \text{ N} - 34.3 \text{ N} = 45.0 \text{ kg} \times a \]\[ 223.2 \text{ N} = 45.0 \text{ kg} \times a \Rightarrow a = \frac{223.2 \text{ N}}{45.0 \text{ kg}} = 4.96 \text{ m/s}^2 \]

Key Concepts

Normal Force Calculation

Normal force is a crucial concept in friction-related problems. It refers to the perpendicular force exerted by a surface to support the weight of an object resting on it. This force is essential for calculating both the static and kinetic friction forces. In this exercise, the normal force is determined by the weight of the crate, which is the product of its mass and gravity. On Earth, this is given by the formula:

• Gravity on Earth: 9.81 m/s²
• Mass of the crate: 45.0 kg

Thus, the normal force \( F_N \) for the crate is calculated as:\[F_N = m \times g = 45.0 \text{ kg} \times 9.81 \text{ m/s}^2 = 441.45 \text{ N}\]This result shows the normal force when the crate is resting on a horizontal floor on Earth, as the surface's reaction balances the weight of the crate.

Static Friction Coefficient

Static friction comes into play when an object is at rest and a force is applied to move it. It is the resistance faced before the object starts to slide. In our crate example, the force needed to start the movement was given as 313 N. The static friction coefficient, denoted by \( \mu_s \), is determined by dividing the static friction force by the normal force.To calculate, use:\[\mu_s = \frac{f_s}{F_N} = \frac{313 \text{ N}}{441.45 \text{ N}} \approx 0.709\]The static friction coefficient measures how strong the interaction is between the crate and the floor, indicating how much force is required to overcome the resistance and initiate motion.

Kinetic Friction Coefficient

Kinetic friction occurs when an object is already in motion. It is usually lower than static friction, as less force is typically needed to keep an object moving than to start the motion. In this situation, a 208 N force keeps the crate sliding steadily across the floor.The kinetic friction coefficient \( \mu_k \) is the ratio of the kinetic friction force to the normal force:\[\mu_k = \frac{f_k}{F_N} = \frac{208 \text{ N}}{441.45 \text{ N}} \approx 0.471\]This calculation tells us how much the floor resists the movement of the crate once it's already in motion, which is crucial for determining how to maintain a constant velocity.

Newton's Second Law

Newton's Second Law of Motion is fundamental for understanding how forces affect the motion of objects. This law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass. The formula is expressed as \( F = m \cdot a \), where \( F \) is the force, \( m \) is the mass of the object, and \( a \) is its acceleration.In the exercise, to find the force to accelerate the crate at \( 1.10 \text{ m/s}^2 \), we consider both the mass of the crate and the kinetic friction:

• Mass: 45.0 kg
• Kinetic friction: 208 N
• Required acceleration: 1.10 m/s²

Thus, the required force is calculated with:\[F = (45.0 \text{ kg} \times 1.10 \text{ m/s}^2) + 208 \text{ N} = 257.5 \text{ N}\]This calculation shows how both friction and desired acceleration influence the amount of force needed.

Gravity on the Moon

Gravity on the Moon is significantly weaker than on Earth, measuring about \( 1.62 \text{ m/s}^2 \) compared to Earth's \( 9.81 \text{ m/s}^2 \). This reduced gravitational pull affects both the normal force and friction forces when conducting the same experiment on the Moon.On the Moon, the normal force \( F_{N,\text{moon}} \) is calculated as:\[F_{N,\text{moon}} = m \times g_{\text{moon}} = 45.0 \text{ kg} \times 1.62 \text{ m/s}^2 = 72.9 \text{ N}\]This lower normal force means less static and kinetic friction. Consequently, the force necessary to initiate movement on the Moon is lower:\[f_{s,\text{moon}} = \mu_s \times F_{N,\text{moon}} = 0.709 \times 72.9 \text{ N} = 51.7 \text{ N}\]This example highlights how different gravitational environments can significantly alter the forces needed to move objects, showcasing the practical implications of physics.