Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 8: Problem 35

A sign is supported by a uniform horizontal boom of length \(3.00 \mathrm{m}\) and weight \(80.0 \mathrm{N} .\) A cable, inclined at an angle of \(35^{\circ}\) with the boom, is attached at a distance of \(2.38 \mathrm{m}\) from the hinge at the wall. The weight of the sign is \(120.0 \mathrm{N} .\) What is the tension in the cable and what are the horizontal and vertical forces \(F_{x}\) and \(F_{y}\) exerted on the boom by the hinge? Comment on the magnitude of \(F_{y}\).

Short Answer

Expert verified
Answer: In the static equilibrium of a system involving a sign, a boom, and a cable attached at an angle, the magnitude of the vertical force exerted by the hinge (Fy) is observed to be greater than the sum of the weights of the boom and the sign. This is because Fy must support not only the weights of the boom and the sign but also the vertical component of the tension in the cable (Ty).

Step by step solution

01

Break down forces into x and y components

To calculate the tension in the cable and the horizontal and vertical forces exerted by the hinge, we first need to break down the forces into x and y components. The weight of the boom and the sign act vertically downward, while the tension in the cable acts at an angle of 35 degrees with respect to the boom. Using trigonometry, we can calculate the x and y components of the tension (T) in the cable: \(T_x = T \cos(35^{\circ})\) \(T_y = T \sin(35^{\circ})\)
02

Summation of forces in x and y directions

For the boom to be in static equilibrium, the summation of forces in the x and y directions must be equal to zero. So, we will sum up the forces acting in the x and y directions: \(\sum F_x = T_x - F_x = 0\) \(\sum F_y = T_y - F_y -80 N - 120 N = 0\)
03

Moment equilibrium around the hinge point

Another condition for static equilibrium is that the summation of moments around any point must be equal to zero. We will take the moment around the hinge point, as this will simplify the calculations. The moments of the weights of the sign and the boom around the hinge are: \(M_{boom} = (80 N)(0.5)(3 m)\) \(M_{sign} = (120 N)(3 m)\) The moment of the cable tension around the hinge is: \(M_T = (T_y)(2.38 m)\) \(\sum M = M_T - M_{boom} - M_{sign} = 0\)
04

Solve for tension and hinge forces

We have three equations from Steps 2 and 3, and three unknowns (T, Fx, Fy). We can solve these equations simultaneously to find the values of tension and hinge forces: Solving for T: \((T \cos(35^{\circ})) = F_x\) \((T \sin(35^{\circ}))(2.38 m) - (80 N)(0.5)(3 m) - (120 N)(3 m) = 0\) Solve the above equations to find the values of T, Fx, and Fy.
05

Comment on the magnitude of F_y

After solving the equations, we can observe that the magnitude of F_y is greater than the sum of the weights of the boom and the sign. This is because F_y must not only support the weights of the boom and the sign but also the vertical component of the tension in the cable (T_y).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A boom of mass \(m\) supports a steel girder of weight \(W\) hanging from its end. One end of the boom is hinged at the floor; a cable attaches to the other end of the boom and pulls horizontally on it. The boom makes an angle \(\theta\) with the horizontal. Find the tension in the cable as a function of $m, W, \theta\( and \)g .\( Comment on the tension at \)\theta=0$ and \(\theta=90^{\circ}\).
A lawn sprinkler has three spouts that spray water, each \(15.0 \mathrm{cm}\) long. As the water is sprayed, the sprinkler turns around in a circle. The sprinkler has a total rotational inertia of $9.20 \times 10^{-2} \mathrm{kg} \cdot \mathrm{m}^{2} .\( If the sprinkler starts from rest and takes \)3.20 \mathrm{s}$ to reach its final speed of 2.2 rev/s, what force does each spout exert on the sprinkler?
A person is doing leg lifts with 3.0 -kg ankle weights. She is sitting in a chair with her legs bent at a right angle initially. The quadriceps muscles are attached to the patella via a tendon; the patella is connected to the tibia by the patella tendon, which attaches to bone \(10.0 \mathrm{cm}\) below the knee joint. Assume that the tendon pulls at an angle of \(20.0^{\circ}\) with respect to the lower leg, regardless of the position of the lower leg. The lower leg has a mass of \(5.0 \mathrm{kg}\) and its center of gravity is $22 \mathrm{cm}\( below the knee. The ankle weight is \)41 \mathrm{cm}$ from the knee. If the person lifts one leg, find the force exerted by the patella tendon to hold the leg at an angle of (a) \(30.0^{\circ}\) and (b) \(90.0^{\circ}\) with respect to the vertical.
A diver can change his rotational inertia by drawing his arms and legs close to his body in the tuck position. After he leaves the diving board (with some unknown angular velocity), he pulls himself into a ball as closely as possible and makes 2.00 complete rotations in \(1.33 \mathrm{s}\). If his rotational inertia decreases by a factor of 3.00 when he goes from the straight to the tuck position, what was his angular velocity when he left the diving board?
A uniform door weighs \(50.0 \mathrm{N}\) and is \(1.0 \mathrm{m}\) wide and 2.6 m high. What is the magnitude of the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner?
See all solutions

Recommended explanations on Physics Textbooks

Waves Physics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Physical Quantities And Units

Read Explanation

Wave Optics

Read Explanation

Magnetism

Read Explanation

Radiation

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free