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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° 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° angle with respect to the vertical. Red's force is along an unknown angle, θ, and has a magnitude of 12 N. Let F₁ be the wolf's force (6.4 N) and F₂ be Red's force (12 N).
02

Decompose the forces into vertical and horizontal components

In order to find the angle θ, we need to first decompose the forces into their vertical and horizontal components. For the wolf's force: - Vertical component: \(F_{1V} = F_1 * cos(25°) = 6.4 * cos(25°)\) - Horizontal component: \(F_{1H} = F_1 * sin(25°) = 6.4 * sin(25°)\) 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: - \((6.4 * cos(25°)) + (12 * cos(\theta)) = F_{Total}\) - \((6.4 * sin(25°)) + (12 * sin(\theta)) = 0\)
04

Solve for the angle θ

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

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Most popular questions from this chapter

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?
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?
A large wooden crate is pushed along a smooth, frictionless surface by a force of \(100 \mathrm{N}\). The acceleration of the crate is measured to be $2.5 \mathrm{m} / \mathrm{s}^{2} .$ What is the mass of the crate?
A skier with a mass of \(63 \mathrm{kg}\) starts from rest and skis down an icy (frictionless) slope that has a length of \(50 \mathrm{m}\) at an angle of \(32^{\circ}\) with respect to the horizontal. At the bottom of the slope, the path levels out and becomes horizontal, the snow becomes less icy, and the skier begins to slow down, coming to rest in a distance of \(140 \mathrm{m}\) along the horizontal path. (a) What is the speed of the skier at the bottom of the slope? (b) What is the coefficient of kinetic friction between the skier and the horizontal surface?
A crow perches on a clothesline midway between two poles. Each end of the rope makes an angle of \(\theta\) below the horizontal where it connects to the pole. If the weight of the crow is \(W,\) what is the tension in the rope? Ignore the weight of the rope.
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