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Chapter 4: Problem 138

A computer weighing \(87 \mathrm{N}\) rests on the horizontal surface of your desk. The coefficient of friction between the computer and the desk is $0.60 .$ (a) Draw an FBD for the computer. (b) What is the magnitude of the frictional force acting on the computer? (c) How hard would you have to push on it to get it to start to slide across the desk?

Short Answer

Expert verified
Answer: A force greater than 52.2 N is needed to make the computer start sliding across the desk.

Step by step solution

01

Draw a Free Body Diagram (FBD)

Sketch a diagram of the computer on the desk. Represent the forces acting on the computer as arrows. There are two forces acting on the computer: the gravitational force (\(F_g\)) pulls the computer downward, and the normal force (\(F_n\)) pushing the computer upwards. Since the computer is sitting on a horizontal surface and not moving vertically, the normal force and gravitational force must be equal and opposite.
02

Calculate the frictional force acting on the computer

We know the normal force (\(F_n\)) is equal to the gravitational force (\(F_g\)), which is given as \(87 N\). The magnitude of the frictional force acting on the computer can be calculated using the formula \(F_f = \mu F_n\), where \(\mu\) is the coefficient of friction and \(F_f\) is the frictional force. Plugging in the given values, we get: \(F_f = 0.60 * 87 \mathrm{N} = 52.2 \mathrm{N}\). The frictional force acting on the computer is \(52.2 \mathrm{N}\).
03

Determine the force needed to make the computer start sliding across the desk

The force needed to start sliding the computer across the desk is the same as the maximum static frictional force since once that force is overcome, the computer will start to move. As we calculated in Step 2, the frictional force is \(52.2 \mathrm{N}\). So, you would have to apply a force greater than \(52.2 \mathrm{N}\) to make the computer start sliding across the desk.

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Most popular questions from this chapter

By what percentage does the weight of an object change when it is moved from the equator at sea level, where the effective value of \(g\) is $9.784 \mathrm{N} / \mathrm{kg},\( to the North Pole where \)g=9.832$ N/kg?
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