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Chapter 11: Problem 59

The left-hand end of a light rod of length \(L\) is attached to a vertical wall by a frictionless hinge. An object of mass \(m\) is suspended from the rod at a point a distance \(\alpha L\) from the hinge, where \(0<\alpha \leq 1.00 .\) The rod is held in a horizontal position by a light wire that runs from the right- hand end of the rod to the wall. The wire makes an angle \(\theta\) with the rod. (a) What is the angle \(\beta\) that the net force exerted by the hinge on the rod makes with the horizontal? (b) What is the value of \(\alpha\) for which \(\beta=\theta ?\) (c) What is \(\beta\) when \(\alpha=1.00 ?\)

Short Answer

Expert verified
The steps involved the construction of free body diagrams, resolving the forces in different directions and applying trigonometric relationships to determine the angles formed by the resultant force at the hinge. The determination of \(\alpha\) required solving equations, while the final task required one to substitute and solve.

Step by step solution

01

Analyzing the Forces

Draw a freebody diagram to understand all the forces acting on the rod. There are three main forces - gravitational force acting at mass m, the force of tension from the wire and the reaction force from the hinge. The gravitational force is acting vertically downward from a point at distance \(\alpha L\) from the hinge. The tension force is acting at an angle \(\theta\) from the point at the end of the rod with the horizontal and the reaction force is acting at the hinge.
02

Calculation of the Resultant Force

By taking the sum of all the forces along the vertical and horizontal directions separately, and equating them to zero, (since in a state of equilibrium, the sum of all forces acting on an object must be zero), we can find the resultant force.
03

Finding \(\beta\)

The angle \(\beta\) that the hinge force makes with the horizontal can be calculated using the resultant horizontal and vertical forces. The tangent of \(\beta\) is the ratio of the vertical resultant force to the horizontal resultant force.
04

Finding \(\alpha\) for \(\beta=\theta\)

For \(\beta=\theta\), equal the expressions for \(\beta\) and \(\theta\) found in steps 3 and 1 respectively. Solve for \(\alpha\). Understanding that \(\alpha\) has to be a value between 0 and 1 can help to refine this solution.
05

Determine \(\beta\) for \(\alpha=1.00\)

Substitute the value \(\alpha=1.00\) into the equation for \(\beta\) found in step 3. Solve for \(\beta\).

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Key Concepts

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

Freebody Diagram
When dealing with problems in equilibrium of forces, especially in physics, creating a freebody diagram can be incredibly useful. A freebody diagram is a visual representation of all the forces acting on an object.
This helps to identify each force's magnitude and direction clearly.
  • Start by sketching the object of interest. In this case, the rod is the primary focus.
  • Mark the forces acting on the object: gravitational force, tension force from the wire, and the hinge reaction force.
  • The gravitational force, exerted by the object's mass, acts vertically downward, placed at a distance of \( \alpha L \) from the hinge.
  • The tension from the wire acts at an angle \( \theta \) from the horizontal at the rod's end.
  • The hinge exerts a reactive force, which may have both horizontal and vertical components.
By visualizing these interactions, a freebody diagram simplifies understanding and solving the problem. It guides the calculation of resultant forces and moments in upcoming steps.
Resultant Force
Once the forces are identified using a freebody diagram, the next step is to calculate the resultant force. The resultant force is essential for determining the equilibrium state of the rod.
In equilibrium, the sum of forces in both the horizontal and vertical directions should be zero. This is because net forces create acceleration, which does not occur in a balanced system.
  • Calculate the sum of forces in the horizontal direction and set it equal to zero.
  • Do the same for the vertical direction.
  • These systems of equations help determine unknown forces or verification of given values.
In this exercise, using the equations of equilibrium, the resultant force on the rod can be balanced.
Once the horizontal and vertical components are known, they assist in calculating the angle \( \beta \). This angle depicts how the net force aligns with the horizontal.
Frictionless Hinge
A frictionless hinge is crucial in this exercise. This hinge allows the rod to rotate without any resistance from friction forces. It simplifies calculations, as one does not need to consider any additional horizontal or vertical forces due to friction.
The hinge only transfers the necessary reactionary forces to maintain the structure in equilibrium.
  • The hinge reaction force can have horizontal and vertical components.
  • These reaction forces are crucial for maintaining balance when force components from tension and gravity are present.
In the context of this problem, the hinge facilitates rotation without altering the overall energy balance in the system.
Being frictionless means mechanical energy is conserved, and these factors enable precise calculations.
Tension Force
Finally, let's explore the concept of tension force as it acts within the wire. Tension force is the pulling force transmitted along a string, cable, or in this problem, a light wire, crucial for maintaining horizontal equilibrium.
  • In this scenario, tension acts at the right-hand end of the rod, making an angle \( \theta \) with it.
  • The tension force balances out a portion of the gravitational force to hold the system stable.
  • Its magnitude and angle significantly affect the resultant force and therefore, the angle \( \beta \).
The concept of tension ensures that parts of the system are not free-falling.
By correctly analyzing the direction and magnitude of the tension, one can solve for unknowns like \( \alpha \) or verify given data points, ensuring the rod remains horizontally aligned.

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Most popular questions from this chapter

A relaxed biceps muscle requires a force of \(25.0 \mathrm{~N}\) for an elongation of \(3.0 \mathrm{~cm} ;\) the same muscle under maximum tension requires a force of \(500 \mathrm{~N}\) for the same elongation. Find Young's modulus for the muscle tissue under each of these conditions if the muscle is assumed to be a uniform cylinder with length \(0.200 \mathrm{~m}\) and cross- sectional area \(50.0 \mathrm{~cm}^{2}\).

Mountaineers often use a rope to lower themselves down the face of a cliff (this is called rappelling). They do this with their body nearly horizontal and their feet pushing against the cliff (Fig. \(\mathbf{P 1 1 . 4 9}\) ). Suppose that an \(82.0 \mathrm{~kg}\) climber, who is \(1.90 \mathrm{~m}\) tall and has a center of gravity \(1.1 \mathrm{~m}\) from his feet, rappels down a vertical cliff with his body raised \(35.0^{\circ}\) above the horizontal. He holds the rope \(1.40 \mathrm{~m}\) from his feet, and it makes a \(25.0^{\circ}\) angle with the cliff face. (a) What tension does his rope need to support? (b) Find the horizontal and vertical components of the force that the cliff face exerts on the climber's feet. (c) What minimum coefficient of static friction is needed to prevent the climber's feet from slipping on the cliff face if he has one foot at a time against the cliff?

A \(15,000 \mathrm{~N}\) crane pivots around a friction-free axle at its base and is supported by a cable making a \(25^{\circ}\) angle with the crane (Fig. E11.20). The crane is \(16 \mathrm{~m}\) long and is not uniform, its center of gravity being \(7.0 \mathrm{~m}\) from the axle as measured along the crane. The cable is attached \(3.0 \mathrm{~m}\) from the upper end of the crane. When the crane is raised to \(55^{\circ}\) above the horizontal holding an \(11,000 \mathrm{~N}\) pallet of bricks by a \(2.2 \mathrm{~m},\) very light cord, find (a) the tension in the cable and (b) the horizontal and vertical components of the force that the axle exerts on the crane. Start with a free-body diagram of the crane.

A uniform, 7.5-m-long beam weighing \(6490 \mathrm{~N}\) is hinged to a wall and supported by a thin cable attached \(1.5 \mathrm{~m}\) 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 uniform drawbridge must be held at a \(37^{\circ}\) angle above the horizontal to allow ships to pass underneath. The drawbridge weighs \(45,000 \mathrm{~N}\) and is \(14.0 \mathrm{~m}\) long. A cable is connected \(3.5 \mathrm{~m}\) from the hinge where the bridge pivots (measured along the bridge) and pulls horizontally on the bridge to hold it in place. (a) What is the tension in the cable? (b) Find the magnitude and direction of the force the hinge exerts on the bridge. (c) If the cable suddenly breaks, what is the magnitude of the angular acceleration of the drawbridge just after the cable breaks? (d) What is the angular speed of the drawbridge as it becomes horizontal?
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