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You are a summer intern for an architectural firm. An 8.00-m-long uniform steel rod is to be attached to a wall by a frictionless hinge at one end. The rod is to be held at 22.0\(^\circ\) below the horizontal by a light cable that is attached to the end of the rod opposite the hinge. The cable makes an angle of 30.0\(^\circ\) with the rod and is attached to the wall at a point above the hinge. The cable will break if its tension exceeds 650 N. (a) For what mass of the rod will the cable break? (b) If the rod has a mass that is 10.0 kg less than the value calculated in part (a), what are the magnitude and direction of the force that the hinge exerts on the rod?

Short Answer

Expert verified
(a) Maximum mass for cable breakage is such that cable tension equals 650 N. (b) Reducing mass by 10 kg, calculate resultant hinge force using component analysis.

Step by step solution

01

Understand the Problem

We have a steel rod attached to a wall with a hinge, held by a cable. We need to find the mass of the rod at which the tension in the cable will reach its breaking point (650 N). Also, if the rod's mass is reduced by 10 kg, we must find the force the hinge exerts on the rod.
02

Calculate the Geometry of the System

The rod is 8.00 meters long, and it's angled 22.0° below horizontal. The cable forms a 30.0° angle with the rod. Using trigonometry, calculate the angles of the forces with respect to the horizontal direction.
03

Set Up Equations for Equilibrium From Torque

For rotational equilibrium, the sum of torques about the hinge must be zero. The torque due to the rod (\( \tau_{rod} = \frac{1}{2} m g L \sin(22°) \)) must equal the torque due to the tension in the cable (\( \tau_{cable} = T \cdot L \cdot \cos(8°) \)) since the angle between the rod and the cable is 30°. Here, L is the length of the rod, m is the mass, and g is acceleration due to gravity.
04

Solve for Mass m

Set the torques equal and solve for the mass of the rod:\[ \frac{1}{2} m g L \sin(22°) = T L \cos(8°) \]Substitute T = 650 N and simplify. Solve for m to find the maximum mass of the rod for which the cable tension is 650 N.
05

Reduce the Mass by 10 kg for Part (b)

Once the maximum mass is found, reduce it by 10 kg to determine the new mass. Use this modified mass to calculate the hinge force in the next steps.
06

Calculate Forces on Rod with New Mass for Part (b)

With the reduced mass, calculate the vertical and horizontal components of the forces acting on the hinge. The vertical component of force is balanced by the weight of the rod and the vertical component of cable tension.
07

Calculate Resultant Hinge Force

Compute the resultant force at the hinge using the Pythagorean theorem \( F_{hinge} = \sqrt{F_{hinge,x}^2 + F_{hinge,y}^2} \). Include direction using trigonometry, calculating angles with respect to the horizontal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Torque
Torque is a crucial concept in physics that describes the rotational equivalent of linear force. It measures how much a force acting on an object causes that object to rotate. In this problem, torque is produced by both the weight of the rod and the tension in the cable.

The formula for calculating torque (\(\tau\)) is given by:\[\tau = r \times F \times \sin(\theta) \]where \(r\) is the position vector (distance from the pivot), \(F\) is the force applied, and \(\theta\) is the angle between the force and the position vector.

To solve these kinds of problems, you must consider the direction and point of application of each force as it can significantly influence the system's rotation.
Rotational Equilibrium
Rotational Equilibrium occurs when the net torque acting on a system is zero, meaning there is no angular acceleration. It is an essential principle for systems where balance and stability are required, like our hinged rod scenario.

For an object to be in rotational equilibrium, the sum of clockwise torques must equal the sum of counterclockwise torques. In the problem, we compare the torque due to the weight of the rod with the torque produced by the tension in the cable.

This ensures that the rod remains static and does not rotate around the hinge. Applying the condition \(\sum \tau = 0\) helps in determining whether the rod will stay steady under the given conditions.
Trigonometry
Trigonometry in physics is used to break down forces into components, particularly when the forces are not aligned with the principal axes. In the case of the steel rod, trigonometry helps calculate angles and components related to the rod's length and inclination.

For example, by knowing the angles the cable and rod make with the horizontal, you can use trigonometric functions such as \(\sin\), \(\cos\), and \(\tan\) to resolve these into perpendicular components.

In this particular setup, \(\cos(8°)\) determines how much of the cable's tension affects the torque about the hinge, and the \(\sin(22°)\) helps calculate the effective perpendicular component of the rod's weight.
Tension
Tension is the pulling force exerted by a string, cable, or similar object on another object. It is a force that tries to extend or elongate the object being pulled, and in our problem, it is the force holding the rod from falling.

For the cable in our scenario, tension must not exceed 650 N, or it will break. By calculating the tension at this point and setting it to 650 N in the equilibrium equation, we determine the critical mass of the rod.

Tension is straightforwardly related to the mass being supported and the angles involved. Thus, understanding these relationships is key in solving problems like determining the threshold at which this tension is exceeded.
Force Analysis
Force analysis involves dissecting all the forces in action in a problem to understand their effects on motion and equilibrium. In the context of our exercise, several forces act on the steel rod: the weight of the rod, the tension from the cable, and the hinge force.

Each of these forces has components along the vertical and horizontal axes that need to be evaluated. For position balance, we calculate both these components and ensure they are counteracted by hinge forces.

By resolving each force into its components, you can find the net force or torque acting on the body and predict its behavior under certain conditions, thereby determining whether the rod under new weight conditions will remain static or move.

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Most popular questions from this chapter

In the human arm, the forearm and hand pivot about the elbow joint. Consider a simplified model in which the biceps muscle is attached to the forearm 3.80 cm from the elbow joint. Assume that the person's hand and forearm together weigh 15.0 N and that their center of gravity is 15.0 cm from the elbow (not quite halfway to the hand). The forearm is held horizontally at a right angle to the upper arm, with the biceps muscle exerting its force perpendicular to the forearm. (a) Draw a free-body diagram for the forearm, and find the force exerted by the biceps when the hand is empty. (b) Now the person holds an 80.0-N weight in his hand, with the forearm still horizontal. Assume that the center of gravity of this weight is 33.0 cm from the elbow. Draw a free-body diagram for the forearm, and find the force now exerted by the biceps. Explain why the biceps muscle needs to be very strong. (c) Under the conditions of part (b), find the magnitude and direction of the force that the elbow joint exerts on the forearm. (d) While holding the 80.0-N weight, the person raises his forearm until it is at an angle of 53.0\(^\circ\) above the horizontal. If the biceps muscle continues to exert its force perpendicular to the forearm, what is this force now? Has the force increased or decreased from its value in part (b)? Explain why this is so, and test your answer by doing this with your own arm.

In a city park a nonuniform wooden beam 4.00 m long is suspended horizontally by a light steel cable at each end. The cable at the left-hand end makes an angle of 30.0\(^\circ\) with the vertical and has tension 620 N. The cable at the right-hand end of the beam makes an angle of 50.0\(^\circ\) with the vertical. As an employee of the Parks and Recreation Department, you are asked to find the weight of the beam and the location of its center of gravity.

A uniform, 7.5-m-long beam weighing 6490 N is hinged to a wall and supported by a thin cable attached 1.5 m from the free end of the beam. The cable runs between the beam and the wall and makes a 40\(^\circ\) angle with the beam. What is the tension in the cable when the beam is at an angle of 30\(^\circ\) above the horizontal?

A 0.120-kg, 50.0-cm-long uniform bar has a small 0.055-kg mass glued to its left end and a small 0.110-kg mass glued to the other end. The two small masses can each be treated as point masses. You want to balance this system horizontally on a fulcrum placed just under its center of gravity. How far from the left end should the fulcrum be placed?

A door 1.00 m wide and 2.00 m high weighs 330 N and is supported by two hinges, one 0.50 m from the top and the other 0.50 m from the bottom. Each hinge supports half the total weight of the door. Assuming that the door's center of gravity is at its center, find the horizontal components of force exerted on the door by each hinge.

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