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Chapter 4: Problem 11

On her way to visit Grandmother, Red Riding Hood sat down to rest and placed her 1.2 -kg basket of goodies beside her. A wolf came along, spotted the basket, and began to pull on the handle with a force of \(6.4 \mathrm{N}\) at an angle of \(25^{\circ}\) with respect to vertical. Red was not going to let go easily, so she pulled on the handle with a force of \(12 \mathrm{N}\). If the net force on the basket is straight up, at what angle was Red Riding Hood pulling?

Short Answer

Expert verified
Answer: Red Riding Hood was pulling at an angle of approximately 20.69掳 below the horizontal.

Step by step solution

01

Identify the forces

Let's first visualize the forces. The wolf's force has a magnitude of 6.4 N and is applied at a 25掳 angle with respect to the vertical. Red's force is along an unknown angle, 胃, and has a magnitude of 12 N. Let F鈧� be the wolf's force (6.4 N) and F鈧� be Red's force (12 N).
02

Decompose the forces into vertical and horizontal components

In order to find the angle 胃, we need to first decompose the forces into their vertical and horizontal components. For the wolf's force: - Vertical component: \(F_{1V} = F_1 * cos(25掳) = 6.4 * cos(25掳)\) - Horizontal component: \(F_{1H} = F_1 * sin(25掳) = 6.4 * sin(25掳)\) For Red's force: - Vertical component: \(F_{2V} = F_2 * cos(\theta) = 12 * cos(\theta)\) - Horizontal component: \(F_{2H} = F_2 * sin(\theta) = 12 * sin(\theta)\)
03

Apply the net force condition

Since the net force on the basket is straight up, the sum of the vertical components of the forces is equal to the total force up, and the sum of the horizontal components is zero. Therefore: - Vertical force: \(F_{1V} + F_{2V} = F_{Total}\) - Horizontal force: \(F_{1H} + F_{2H} = 0\) Substitute the expressions for vertical and horizontal components calculated earlier: - \((6.4 * cos(25掳)) + (12 * cos(\theta)) = F_{Total}\) - \((6.4 * sin(25掳)) + (12 * sin(\theta)) = 0\)
04

Solve for the angle 胃

From the horizontal force equation, we can solve for the angle 胃: \(12 * sin(\theta) = -6.4 * sin(25掳)\) \(\theta = arcsin(-\frac{6.4 * sin(25掳)}{12})\) Now calculate the angle: \(\theta = arcsin(-\frac{6.4 * sin(25掳)}{12}) \approx -20.69掳\) Since the angle is negative, it means that Red Riding Hood was pulling at an angle of 20.69掳 below the horizontal.

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Most popular questions from this chapter

An \(80.0-\mathrm{N}\) crate of apples sits at rest on a ramp that runs from the ground to the bed of a truck. The ramp is inclined at \(20.0^{\circ}\) to the ground. (a) What is the normal force exerted on the crate by the ramp? (b) The interaction partner of this normal force has what magnitude and direction? It is exerted by what object on what object? Is it a contact or a long-range force? (c) What is the static frictional force exerted on the crate by the ramp? (d) What is the minimum possible value of the coefficient of static friction? (e) The normal and frictional forces are perpendicular components of the contact force exerted on the crate by the ramp. Find the magnitude and direction of the contact force.
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 book rests on the surface of the table. Consider the following four forces that arise in this situation: (a) the force of the Earth pulling on the book, (b) the force of the table pushing on the book, (c) the force of the book pushing on the table, and (d) the force of the book pulling on the Earth. The book is not moving. Which pair of forces must be equal in magnitude and opposite in direction even though they are not an interaction pair?
Margie, who weighs \(543 \mathrm{N},\) is standing on a bathroom scale that weighs \(45 \mathrm{N}\). (a) With what force does the scale push up on Margie? (b) What is the interaction partner of that force? (c) With what force does the Earth push up on the scale? (d) Identify the interaction partner of that force.
The coefficient of static friction between a block and a horizontal floor is \(0.40,\) while the coefficient of kinetic friction is \(0.15 .\) The mass of the block is \(5.0 \mathrm{kg} .\) A horizontal force is applied to the block and slowly increased. (a) What is the value of the applied horizontal force at the instant that the block starts to slide? (b) What is the net force on the block after it starts to slide?
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