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Chapter 4: Problem 151

In a movie, a stuntman places himself on the front of a truck as the truck accelerates. The coefficient of friction between the stuntman and the truck is \(0.65 .\) The stuntman is not standing on anything but can "stick" to the front of the truck as long as the truck continues to accelerate. What minimum forward acceleration will keep the stuntman on the front of the truck?

Short Answer

Expert verified
Answer: The minimum forward acceleration required is approximately 8.175 m/s虏.

Step by step solution

01

Identify the forces acting on the stuntman

We can start by identifying the forces acting on the stuntman. There are two main forces: 1. Gravitational force (Fg) acting downwards. 2. Frictional force (Ff) acting upwards along the front of the truck, which opposes the gravitational force.
02

Apply Newton's second law of motion

According to Newton's second law of motion, the net force acting on an object is equal to the product of the object's mass (m) and acceleration (a). In our case, the net force acting on the stuntman is the difference between the frictional force and the gravitational force: Net force (F) = Ff - Fg
03

Calculate the frictional force

To find the frictional force acting on the stuntman, we'll use the equation: Ff = 渭 * Fn Where 渭 is the coefficient of friction (0.65) and Fn is the normal force acting on the stuntman, which is equal to Fg (m * g) since there is no vertical acceleration.
04

Calculate the net force

Now, we can rewrite the net force equation as: F = 渭 * Fg - Fg
05

Find the minimum acceleration to keep the stuntman on the truck

We know that F = m * a, and we want to find the acceleration (a). We can rewrite the equation in terms of acceleration: a = F / m Substitute the net force equation from Step 4: a = (渭 * Fg - Fg) / m Since Fg = m * g, we can replace Fg with m * g: a = (渭 * m * g - m * g) / m Cancel out the mass (m) from the equation: a = (渭 * g - g) Now, plug in the values for 渭 (0.65) and g (9.81 m/s虏): a = (0.65 * 9.81 - 9.81) a = 1.635 - 9.81 a 鈮� -8.175 m/s虏 However, the negative sign indicates that the stuntman is actually accelerating backward relative to the front of the truck. He's staying in place relative to the ground, but the truck is moving forward. So, we can take the absolute value of the acceleration: a = 8.175 m/s虏 (approx.)
06

State the answer

The minimum forward acceleration required to keep the stuntman on the front of the truck is approximately 8.175 m/s虏.

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Most popular questions from this chapter

A 2.0 -kg toy locomotive is pulling a 1.0 -kg caboose. The frictional force of the track on the caboose is \(0.50 \mathrm{N}\) backward along the track. If the train's acceleration forward is \(3.0 \mathrm{m} / \mathrm{s}^{2},\) what is the magnitude of the force exerted by the locomotive on the caboose?
You grab a book and give it a quick push across the top of a horizontal table. After a short push, the book slides across the table, and because of friction, comes to a stop. (a) Draw an FBD of the book while you are pushing it. (b) Draw an FBD of the book after you have stopped pushing it, while it is sliding across the table. (c) Draw an FBD of the book after it has stopped sliding. (d) In which of the preceding cases is the net force on the book not equal to zero? (e) If the book has a mass of \(0.50 \mathrm{kg}\) and the coefficient of friction between the book and the table is \(0.40,\) what is the net force acting on the book in part (b)? (f) If there were no friction between the table and the book, what would the free-body diagram for part (b) look like? Would the book slow down in this case? Why or why not?
A sailboat, tied to a mooring with a line, weighs \(820 \mathrm{N}\) The mooring line pulls horizontally toward the west on the sailboat with a force of $110 \mathrm{N}$. The sails are stowed away and the wind blows from the west. The boat is moored on a still lake-no water currents push on it. Draw an FBD for the sailboat and indicate the magnitude of each force.
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A box full of books rests on a wooden floor. The normal force the floor exerts on the box is \(250 \mathrm{N}\). (a) You push horizontally on the box with a force of \(120 \mathrm{N},\) but it refuses to budge. What can you say about the coefficient of static friction between the box and the floor? (b) If you must push horizontally on the box with a force of at least \(150 \mathrm{N}\) to start it sliding, what is the coefficient of static friction? (c) Once the box is sliding, you only have to push with a force of \(120 \mathrm{N}\) to keep it sliding. What is the coefficient of kinetic friction?
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