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Chapter 5: Problem 29

A 45.0-kg crate of tools rests on a horizontal floor. You exert a gradually increasing horizontal push on it, and the crate just begins to move when your force exceeds 313 N. Then you must reduce your push to 208 N to keep it moving at a steady 25.0 cm/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 m/s\(^2\)? (c) Suppose you were performing the same experiment on the moon, where the acceleration due to gravity is 1.62 m/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)?

Short Answer

Expert verified
(a) \( \mu_s = 0.707 \), \( \mu_k = 0.472 \). (b) F = 257 N. (c) (i) 51.2 N, (ii) 5.26 m/s虏.

Step by step solution

01

Understanding Static Friction

To find the coefficient of static friction \( \mu_s \), we use the maximum static friction force to overcome before the crate starts moving. This is given by \( f_s = \mu_s \times F_n \), where \( F_n \), the normal force, is equal to the weight of the crate since the floor is horizontal \( F_n = mg \). Thus, \( 313 \text{ N} = \mu_s \times (45.0 \text{ kg} \times 9.81 \text{ m/s}^2) \). Calculate \( \mu_s \).
02

Understanding Kinetic Friction

Once the crate is moving, the kinetic frictional force can be calculated using \( f_k = \mu_k \times F_n \). Since the force required to keep the crate moving at a constant velocity is 208 N, we use gravitational force \( F_n = mg = 45.0 \text{ kg} \times 9.81 \text{ m/s}^2 \) to find \( \mu_k \). Thus, \( 208 \text{ N} = \mu_k \times (45.0 \text{ kg} \times 9.81 \text{ m/s}^2) \). Calculate \( \mu_k \).
03

Push for Acceleration

To achieve an acceleration \( a = 1.10 \text{ m/s}^2 \), apply Newton's Second Law: the net force needed is \( F_{net} = ma + f_k \), where \( f_k = \mu_k \times mg \) as calculated before. Calculate the total force \( F_{apply} = F_{net} + f_k \) required.
04

Static Friction on the Moon

The gravitational acceleration on the moon is \( g_{moon} = 1.62 \text{ m/s}^2 \). Recalculate the normal force \( F_{n,moon} = 45.0 \text{ kg} \times 1.62 \text{ m/s}^2 \) to find the static friction force \( f_{s,moon} = \mu_s \times F_{n,moon} \). Use earlier static friction coefficient \( \mu_s \) to find the new static force.
05

Acceleration on the Moon

Using \( F_{apply} \) from the earlier calculation and the moon's lower gravitational force, determine the new net force and acceleration \( a_{moon} \). Recalculate using \( F_{net,moon} = F_{apply} - \mu_k \times F_{n,moon} \) to find the new acceleration \( a_{moon} = \frac{F_{net,moon}}{m} \).

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Key Concepts

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

Static Friction
Static friction is the force that keeps an object at rest when a force is applied. It must be overcome for the object to start moving. In the context of our exercise, static friction is what holds the crate steady. Before it begins moving, the force exerted (313 N) exactly balances the static frictional force.
To find the coefficient of static friction \(\mu_s\), use the formula:
  • \(f_s = \mu_s \times F_n\)
  • where \(F_n = mg\) (the weight of the crate since the surface is horizontal)
Replace the forces to get \(313 \text{ N} = \mu_s \times (45.0 \text{ kg} \times 9.81 \text{ m/s}^2)\).
Solving for \(\mu_s\), you find it indicates the stickiness between surfaces. In practical terms, it tells you how hard you need to push to break the crate free from rest.
Kinetic Friction
Once the object is in motion, static friction is replaced by kinetic friction, which is usually lower in value. In this case, the moving crate has a steady velocity when pushed with 208 N, which highlights the influence of kinetic friction.
The kinetic frictional force can be calculated as follows:
  • \(f_k = \mu_k \times F_n\)
  • Using the same \(F_n = mg\), since gravity hasn't changed
Now, 208 N keeps the crate moving, so \(208 \text{ N} = \mu_k \times (45.0 \text{ kg} \times 9.81 \text{ m/s}^2)\).
This will help you calculate \(\mu_k\), the coefficient of kinetic friction, crucial for understanding how readily an object moves once it's been nudged into motion. It highlights ease of sliding once motion begins, often less than the force needed to start it.
Normal Force
The normal force is the support force exerted perpendicular to the surface. It's essential for calculating friction because it acts as the weight's counterbalance.
For a horizontal surface, this is simplified to the gravitational force, \(F_n = mg\). This makes calculations easier as you multiply the mass by gravitational acceleration:
  • On Earth: \(F_n = 45.0 \text{ kg} \times 9.81 \text{ m/s}^2\)
  • On the Moon: the calculation changes due to \(g_{moon} = 1.62 \text{ m/s}^2\)
Understanding the normal force's role helps visualize how weight influences friction. It's more than just an upward force; it's the baseline for finding frictional forces which are essential for any movement or resistance analysis.
Acceleration
Acceleration is the rate of change of velocity per unit time. To change an object's speed, you need to overcome friction and produce a new net force.
In our exercise, to make the crate accelerate at \(1.10 \text{ m/s}^2\), apply Newton's second law. This relates acceleration, force, and mass:
  • Net Force, \(F_{net} = ma + f_k\)
  • Total applied force, \(F_{apply} = F_{net} + f_k\)
So use this net force in calculations to find the needed push for extra speed.
Even on the Moon, with different gravity, this principle holds. Just account for lower gravitational pull, leading to new net forces and different acceleration. It illustrates how changing environments affect force and acceleration dynamics.

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Most popular questions from this chapter

A 2540-kg test rocket is launched vertically from the launch pad. Its fuel (of negligible mass) provides a thrust force such that its vertical velocity as a function of time is given by \(v(t) = At + Bt^2\), where \(A\) and \(B\) are constants and time is measured from the instant the fuel is ignited. The rocket has an upward acceleration of 1.50 m/s\(^2\) at the instant of ignition and, 1.00 s later, an upward velocity of 2.00 m>s. (a) Determine \(A\) and \(B\), including their SI units. (b) At 4.00 s after fuel ignition, what is the acceleration of the rocket, and (c) what thrust force does the burning fuel exert on it, assuming no air resistance? Express the thrust in newtons and as a multiple of the rocket's weight. (d) What was the initial thrust due to the fuel?

An airplane flies in a loop (a circular path in a vertical plane) of radius 150 m. The pilot's head always points toward the center of the loop. The speed of the airplane is not constant; the airplane goes slowest at the top of the loop and fastest at the bottom. (a) What is the speed of the airplane at the top of the loop, where the pilot feels weightless? (b) What is the apparent weight of the pilot at the bottom of the loop, where the speed of the airplane is 280 km/h? His true weight is 700 N.

A 1125-kg car and a 2250-kg pickup truck approach a curve on a highway that has a radius of 225 m. (a) At what angle should the highway engineer bank this curve so that vehicles traveling at 65.0 mi/h can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? (b) As the car and truck round the curve at 65.0 mi/h, find the normal force on each one due to the highway surface.

Block \(A\), with weight \(3w\), slides down an inclined plane \(S\) of slope angle 36.9\(^{\circ}\) at a constant speed while plank \(B\), with weight \(w\), rests on top of A. The plank is attached by a cord to the wall \(\textbf{(Fig. P5.97).}\) (a) Draw a diagram of all the forces acting on block \(A\). (b) If the coefficient of kinetic friction is the same between \(A\) and \(B\) and between \(S\) and \(A\), determine its value.

A rocket of initial mass 125 kg (including all the contents) has an engine that produces a constant vertical force (the \(thrust\)) of 1720 N. Inside this rocket, a 15.5-N electrical power supply rests on the floor. (a) Find the initial acceleration of the rocket. (b) When the rocket initially accelerates, how hard does the floor push on the power supply? (\(Hint:\) Start with a free- body diagram for the power supply.)
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