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A worker pushes horizontally on a \(35 \mathrm{~kg}\) crate with a force of magnitude \(110 \mathrm{~N}\). The coefficient of static friction between the crate and the floor is 0.37 . (a) What is the value of \(f_{s, \max }\) under the circumstances? (b) Does the crate move? (c) What is the frictional force on the crate from the floor? (d) Suppose, next, that a second worker pulls directly upward on the crate to help out. What is the least vertical pull that will allow the first worker's \(110 \mathrm{~N}\) push to move the crate? (e) If, instead, the second worker pulls horizontally to help out, what is the least pull that will get the crate moving?

Short Answer

Expert verified
(a) 126.91 N, (b) No, (c) 110 N, (d) 14.19 N, (e) 16.91 N.

Step by step solution

01

Calculate Maximum Static Frictional Force

First, determine the maximum static frictional force (\(f_{s,\text{max}}\)) using the formula: \[f_{s,\text{max}} = \mu_s \cdot F_n\] where \(\mu_s = 0.37\) is the coefficient of static friction and \(F_n\) is the normal force. The normal force \(F_n\) is equal to the weight of the crate, which is \(35 \times 9.8 = 343\, \text{N}\). Substituting the values into the equation gives \[f_{s,\text{max}} = 0.37 \times 343 = 126.91\, \text{N}.\]
02

Determine if the Crate Moves

The crate will move if the pushing force exceeds the maximum static frictional force. The worker applies a force of \(110\, \text{N}\). Since \(110\, \text{N} < 126.91\, \text{N}\), the crate does not move.
03

Calculate Actual Frictional Force

If the crate doesn't move, the static frictional force is equal to the applied force. Thus, the frictional force on the crate is \(110\, \text{N}\).
04

Find Minimum Vertical Pull to Move the Crate

With a second worker pulling upward, the normal force decreases, which reduces static friction. Let \(F_{n}' = F_n - F_{\text{pull}}\) be the new normal force. The reduced frictional force needed for movement is \(f_{s,\text{max}}' = \mu_s \cdot F_{n}'\) and should be less than \(110\, \text{N}\). Solve \[110 = 0.37 \cdot (343 - F_{\text{pull}})\] to find \(F_{\text{pull}} = 14.19\, \text{N}.\)
05

Find Minimum Horizontal Pull to Move the Crate

If the second worker pulls horizontally, the total horizontal force is \(110 + F_{\text{pull}}\). The total force must exceed \(126.91\, \text{N}\). Solving \(110 + F_{\text{pull}} > 126.91\) gives \(F_{\text{pull}} = 16.91\, \text{N}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Force
Normal force is a vital concept in physics, describing the force perpendicular to a surface. It is exerted by a surface to support the weight of an object resting upon it. This force is crucial because it balances the object's weight when stable on a surface, preventing it from falling through. In the context of the exercise, the normal force is the force exerted by the floor on the 35 kg crate. To calculate the normal force, we need to consider the weight of the crate. The force due to gravity on the crate is calculated using the formula: \[ F_n = m \cdot g \] where \( m \) is the mass (35 kg) and \( g \) is the acceleration due to gravity (approximately \(9.8\, \text{m/s}^2\)). Plugging in these values, we find that the normal force \( F_n \) equals \(343\, \text{N}\). This normal force plays a pivotal role in determining the frictional forces at work.
Static Frictional Force
Static frictional force is the force that acts to resist the initiation of sliding motion between two surfaces in contact. It is the reason objects do not slide down a sloped surface or remain still when a small force is applied. The formula to determine the static frictional force \(f_{s,\text{max}}\) is: \[ f_{s,\text{max}} = \mu_s \cdot F_n \] where \( \mu_s \) is the coefficient of static friction, and \( F_n \) is the normal force. In our crate exercise, with \( \mu_s = 0.37 \) and \( F_n = 343 \text{ N} \), the maximum static frictional force is found to be \( 126.91 \text{ N} \). This value is the threshold force required to overcome static friction and initiate the crate's movement. When the applied force is less than \(f_{s,\text{max}}\), as it is in this case (\( 110 \text{ N}\)), the static frictional force equals the applied force (\( 110 \text{ N}\)), preventing motion.
Newton's Laws of Motion
Newton's laws of motion form the foundation for understanding mechanics in physics. These laws describe the relationship between a body and the forces acting on it, and how it moves as a result.1. **First Law**: An object will remain at rest or in uniform motion unless acted upon by an external force. In this exercise, the crate does not move because the static frictional force is equal to the applied force, illustrating the first law. 2. **Second Law**: The acceleration of an object depends on the net force acting on it and its mass, expressed as \( F = m \cdot a \). If the net force exceeds the static frictional force, the crate would accelerate. 3. **Third Law**: For every action, there is an equal and opposite reaction. When the worker pushes the crate, the floor pushes back with an equal force of static friction.Understanding these laws empowers us to predict an object's motion or the forces necessary to alter that motion. They are critical for comprehending why the crate remains stationary or what additional forces are needed to induce movement.

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Most popular questions from this chapter

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