/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 39 A dockworker loading crates on a... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 39

A dockworker loading crates on a ship finds that a \(20-\mathrm{kg}\) crate, initially at rest on a horizontal surface, requires a \(75-\mathrm{N}\) horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of \(60 \mathrm{~N}\) is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor.

Short Answer

Expert verified
The coefficient of static friction between the crate and the floor is approx. 0.38 and the coefficient of kinetic friction is approx. 0.31.

Step by step solution

01

Determine Normal Force

The normal force (\(N\)) on the crate is the weight of the crate. It can be determined by multiplying mass of the crate with the acceleration due to gravity (\(g\)). In this case, \(N = 20 \, \text{kg} \times 9.8 \, \text{m/s}^{2} = 196 \, \text{N}\).
02

Find Coefficient of Static Friction

The coefficient of static friction (\(\mu_s\)) is found by dividing the force required to get the crate moving (the force of static friction) by the normal force. So, \(\mu_s = \frac{f_s}{N} = \frac{75 \, \text{N}}{196 \, \text{N}} = 0.38\), rounded to two decimal places.
03

Find Coefficient of Kinetic Friction

The coefficient of kinetic friction (\(\mu_k\)) is determined by dividing the force required to keep the crate moving (the force of kinetic friction) by the normal force. So, \(\mu_k = \frac{f_k}{N} = \frac{60 \, \text{N}}{196 \, \text{N}} = 0.31\), rounded to two decimal places.

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Most popular questions from this chapter

A hockey puck struck by a hockey stick is given an initial speed \(v_{0}\) in the positive \(x\)-direction. The coefficient of kinetic friction between the ice and the puck is \(\mu_{k^{*}}\) (a) Obtain an expression for the acceleration of the puck. (b) Use the result of part (a) to obtain an expression for the distance \(d\) the puck slides. The answer should be in terms of the variables \(v_{0}, \mu_{k}\), and \(g\) only.

A crate of weight \(F_{g}\) is pushed by a force \(\overrightarrow{\mathbf{P}}\) on a horizontal floor as shown in Figure \(\mathrm{P} 4.83\). The coefficient of static friction is \(\mu_{s}\), and \(\overrightarrow{\mathbf{P}}\) is directed at angle \(\theta\) below the horizontal. (a) Show that the minimum value of \(P\) that will move the crate is given by $$ P=\frac{\mu_{s} F_{g} \sec \theta}{1-\mu_{s} \tan \theta} $$ (b) Find the condition on \(\theta\) in terms of \(\mu_{s}\) for which motion of the crate is impossible for any value of \(P\).

The parachute on a race car of weight \(8820 \mathrm{~N}\) opens at the end of a quarter-mile run when the car is traveling at \(35 \mathrm{~m} / \mathrm{s}\). What total retarding force must be supplied by the parachute to stop the car in a distance of \(1000 \mathrm{~m}\) ?

As a protest against the umpire's calls, a baseball pitcher throws a ball straight up into the air at a speed of \(20.0 \mathrm{~m} / \mathrm{s}\). In the process, he moves his hand through a distance of \(1.50 \mathrm{~m}\). If the ball has a mass of \(0.150 \mathrm{~kg}\), find the force he exerts on the ball to give it this upward speed.

The force exerted by the wind on the sails of a sailboat is \(390 \mathrm{~N}\) north. The water exerts a force of \(180 \mathrm{~N}\) east. If the boat (including its crew) has a mass of \(270 \mathrm{~kg}\), what are the magnitude and direction of its acceleration?
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