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Chapter 11: Problem 21

A 3.00-m-long, \(190 \mathrm{~N}\), uniform rod at the \(\mathrm{zoo}\) is held in a horizontal position by two ropes at its ends (Fig. E11.21). The left rope makes an angle of \(150^{\circ}\) with the rod, and the right rope makes an angle \(\theta\) with the horizontal. A \(90 \mathrm{~N}\) howler monkey (Alouatta seniculus) hangs motionless \(0.50 \mathrm{~m}\) from the right end of the rod as he carefully studies you. Calculate the tensions in the two ropes and the angle \(\theta\). First make a free-body diagram of the rod.

Short Answer

Expert verified
To solve this exercise, we first need to calculate the total weight of the rod and the monkey. After that, we analyze the forces and apply the conditions of equilibrium to set up equations for the tensions in the two ropes. Solving these equations will allow us to find the tensions in the ropes and the angle 胃. The answer will be numerical values of T1, T2 and 胃.

Step by step solution

01

Calculate the total weight

The total weight acting on the centre of the rod can be calculated as the sum of weights of the rod and monkey. Use the equation \(W_{total}=W_{rod}+W_{monkey}\), where \(W_{rod}=mg=190N\) and \(W_{monkey}=mg=90N\). So, \(W_{total}=190N+90N=280N\).
02

Analyze forces

There are three forces acting on the rod: The tension T1 in the left rope, the tension T2 in the right rope and the total weight of the rod and the monkey, acting downwards. These forces should sum up to zero because of Newton's second law.
03

Apply conditions of equilibrium

Since the rod is in equilibrium, the sum of torques acting on it should be zero. Setting torques with respect to the left end of the rod (since T1, being at the pivot point, will not contribute to the torque), we write down the torque due to T2 vertically upwards and the total weight W acting downwards. The torque due to T2 is \(T2*cos(\theta)*3m\), and the torque due to the total weight W is \(280N*1.5m\). Setting these equal gives us the equation \(T2*cos(\theta)*3m=280N*1.5m\).
04

Solve for T2

Solve the equation from step 3 to find the tension T2 in the right rope.
05

Find T1 and theta

Use the force balance in the vertical direction (sum of forces is zero) to find the tension T1 in the left rope. Once we know T1 and T2, we can determine the angle 胃. T1 is equal to \(T2 * cos(胃) - W\) and 胃 is equal to \(arccos(T1 / T2) + W\). Solving these equations will give us T1 and 胃.

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Key Concepts

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

Free-body Diagram
Understanding a free-body diagram is crucial when analyzing problems involving forces and equilibrium. In this scenario, we have a uniform rod suspended by two ropes and subjected to gravitational forces. To draw a free-body diagram, follow these simple steps:

  • Identify the object: Here, it is the rod.
  • Show all the forces: Draw the weight of the rod and the monkey acting downwards. Represent their sum as a single force acting at the center of mass of the system.
  • Illustrate the tensions: Show the tension forces in each rope. Tension T1 acts via the left rope, while tension T2 comes from the right rope.
  • Indicate the angles: Mark the angle between the left rope and the rod (150掳) and the corresponding angle 胃 with the horizontal for the right rope.
A clear diagram helps isolate all forces acting on the object, making it easier to apply further analytical techniques to solve for unknowns.
Newton's Second Law
Newton's Second Law provides a fundamental principle used in equilibrium problems: for an object at rest, the forces are balanced. The law is expressed as:

\[ \Sigma F = ma \]

For an object in static equilibrium, the acceleration \(a\) is zero, simplifying the equation to:

\[ \Sigma F = 0 \]

In the exercise, three main forces act on the rod:
  • The gravitational force, which is the sum of the weights of the rod and monkey (280 N downward).
  • The tension T1 from the left rope, which contributes a force, often resolved into vertical and horizontal components.
  • The tension T2 from the right rope, similarly broken down into components, helps counterbalance the other forces.
By setting up equations for the sum of vertical and horizontal forces to be zero, we ensure the rod remains in its horizontal position without accelerating.
Conditions of Equilibrium
Conditions of equilibrium ensure that an object remains stable and unmoving. This involves balancing both linear forces and rotational torques. The conditions are:

  • Translational equilibrium: The sum of all forces in any direction must be zero.
  • Rotational equilibrium: The sum of all torques about any axis must be zero.

In the rod exercise, translational equilibrium is used to equilibrate the vertical and horizontal forces from the two tensions and the weights.

Rotational equilibrium focuses on torques, determined by both force magnitude and the distance at which they act (lever arm). Using the left end of the rod as a pivot, the counteracting torques from tension T2 and the total weight ensure the rod鈥檚 balanced state:

\[ T2 \times \cos(\theta) \times 3 = 280 \times 1.5 \]

By applying these conditions effectively, both tensions and the unknown angle 胃 can be solved, maintaining the rod's equilibrium.

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Most popular questions from this chapter

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 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?

In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of \(90.8 \mathrm{~N}\) is applied perpendicular to each end. If the diameter of the wire is \(1.84 \mathrm{~mm},\) what is the breaking stress of the alloy?

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 \(50.0 \mathrm{~cm}^{2}\).

A door \(1.00 \mathrm{~m}\) wide and \(2.00 \mathrm{~m}\) high weighs \(330 \mathrm{~N}\) and is supported by two hinges, one \(0.50 \mathrm{~m}\) from the top and the other \(0.50 \mathrm{~m}\) 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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