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\(\mathrm{A} 3.00 \mathrm{-m}-\mathrm{long}, 240-\mathrm{N},\) uniform rod at the zoo is held in a horizontal position by two ropes at its ends (Fig. Ell. 19). The left rope makes an angle of \(150^{\circ}\) with the rod and the right rope makes an angle \(\theta\) with the horizontal. \(\mathrm{A} 90\) -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
T鈧 鈮 247 N, T鈧 鈮 143 N, 胃 鈮 43.9掳.

Step by step solution

01

Understand the Scenario

Visualize the rod held horizontally by two ropes, with a monkey hanging 0.50 meters from the right end. The left rope is pulling upward and to the left at an angle of 150掳, and the right rope is pulling upwards at an unknown angle 胃.
02

Draw the Free-body Diagram

Draw a diagram with forces acting on the rod: the weight of the rod acting downwards at its center, the weight of the monkey acting downwards 0.50 meters from the right end, and the tensions T鈧 in the left rope and T鈧 in the right rope.
03

Apply Equilibrium Conditions

For the rod to be in equilibrium, the sum of all forces and the sum of all torques acting on it must be zero. This provides equations to calculate tensions and the angle.
04

Set Up Force Equations

The vertical forces must cancel each other out:\[T_1 \sin 150掳 + T_2 \sin 胃 = 240 \, \text{N} + 90 \, \text{N}\]This equation ensures the vertical forces are balanced.
05

Set Up Torque Equation

Select a pivot point, commonly at one end of the rod. Calculate torques about that point to find:\[T_1 (3.00) \sin 30掳 = 90 \, \text{N} (0.50) + 240 \, \text{N} (1.50)\]Solve for T鈧.
06

Solve for T鈧 and T鈧

The equation in Step 5 determines T鈧. Use this T鈧 value and the force equations to solve for T鈧 using:\[T_2 \sin 胃 = (240 \, \text{N} + 90 \, \text{N}) - T_1 \sin 150掳\]
07

Solve for Angle 胃

Using T鈧 from the previous step, calculate 胃 by rearranging the force equation for horizontal components:\[T_2 \cos 胃 = T_1 \cos 150掳\]Solve for 胃.

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Key Concepts

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

Free-body Diagram
A free-body diagram is a crucial step when analyzing the equilibrium of rigid bodies. It visually represents all the forces acting on the object, in this case, the rod at the zoo. Imagine the rod as a horizontal line in the diagram. Next, depict each force:
  • The weight of the 240-N rod acting downward at its center.
  • The weight of the 90-N howler monkey, pulling downwards 0.50 m from the right end.
  • Tension force T鈧, pulling upwards and to the left on the left end of the rod, making an angle of 150掳 with the rod.
  • Tension force T鈧, pulling upwards at an unknown angle 胃 on the right end of the rod.
By associating each force graphically, the free-body diagram simplifies identifying the forces involved and setting up relevant equilibrium equations. It's a vital step that helps visualize how the forces interact to keep the system balanced.
Torque
Torque is the rotational equivalent of linear force, describing how effectively a force causes an object to rotate. When analyzing the rod, torque will determine how balanced it is around a pivot point. Think of the rod as a seesaw; torque is what keeps it from tipping over. In equilibrium scenarios, the sum of all torques must equal zero. We calculate torque using the formula:\[\text{Torque} = \text{Force} \times \text{Distance} \times \sin(\text{angle})\]Select a pivot point, like the left or right end of the rod. Computing torques for all forces relative to this point allows determining tension in the ropes:
  • Torque due to T鈧 involves the entire length of the rod since it acts at the opposite end from the pivot.
  • The torque due to the monkey's weight acts at its position 0.50 m from the right end.
  • The torque due to the rod鈥檚 weight acts at its midpoint, 1.5 m from either end.
Setting the sum of these torques to zero is vital for maintaining equilibrium.
Force Components
Force components break down a force into vertical and horizontal parts. They are essential when dealing with angled forces, like the tensions in the ropes. Visualizing force components helps tackle complex problems by simplifying them into effective forces in axes perpendicular to each other:
  • For the left rope at 150掳, the tension T鈧 divides into horizontal (T鈧乧os150掳) and vertical (T鈧乻in150掳) components.
  • For the right rope making angle 胃, the tension T鈧 has components: horizontal (T鈧俢os胃) and vertical (T鈧俿in胃).
To ensure the rod is stationary, sum total vertical and horizontal forces must be zero. The vertical equation balances the weight of the rod and monkey with the vertical components of T鈧 and T鈧. Similarly, the horizontal equation ensures horizontal components cancel out. This breakdown into components is crucial for deriving the necessary equations to solve for unknowns like tensions and angles.
Tension
Tension refers to the pulling force transmitted along a string, rope, or similar object when the object is parallel to the force being applied. In this problem, tensions T鈧 and T鈧 keep the rod in horizontal equilibrium. Both act upward and oppose the gravitational pull on the rod and monkey. Imagine stretching a rubber band; the tension keeps it taut. Here鈥檚 why tensions matter and how they're calculated:
  • For T鈧, the vertical component (T鈧乻in150掳) aids in counteracting the downward forces.
  • For T鈧, its vertical component (T鈧俿in胃) works to offset gravity.
Lastly, determining the tensions involves setting the algebraic sum of forces and torques to zero, leveraging calculated force components. Knowing how to calculate tensions in a system is key to solving equilibrium problems effectively, ensuring that all parts of a system are in steady, balanced states.

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Most popular questions from this chapter

Suppose that you can lift no more than 650 \(\mathrm{N}\) (around 150 lb) unaided. (a) How much can you lift using a \(1.4-\mathrm{m}\) -long wheelbarrow that weighs 80.0 \(\mathrm{N}\) and whose center of gravity is 0.50 \(\mathrm{m}\) from the center of the wheel (Fig. E11.16)? The center of gravity of the load carried in the wheelbarrow is also 0.50 \(\mathrm{m}\) from the center of the wheel. (b) Where does the force come from to enable you to lift more than 650 \(\mathrm{N}\) using the wheelbarrow?

A uniform rod is 2.00 \(\mathrm{m}\) long and has 1.80 \(\mathrm{kg} . \mathrm{A}\) \(2.40-\) -kg clamp is attached to the rod. How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.20 \(\mathrm{m}\) from the left-hand end of the rod?

An engineer is designing a conveyor system for loading hay bales into a wagon (Fig. P11.81). Each bale is 0.25 \(\mathrm{m}\) wide, 0.50 \(\mathrm{m}\) high, and 0.80 \(\mathrm{m}\) long (the dimension perpendicular to the plane of the figure), with mass 30.0 kg. The center of gravity of each bale is at its geometrical center. The coefficient of static friction between a bale and the conveyor belt is 0.60 , and the belt moves with constant speed. (a) The angle \(\beta\) of the conveyor is slowly increased. At some critical angle a bale will tip (if it doesn't slip first), and at some different critical angle it will slip (if it doesn't tip first). Find the two critical angles and determine which happens at the smaller angle. (b) Would the outcome of part (a) be different if the coefficient of friction were 0.40\(?\)

A 350 -N, uniform, 1.50 -m bar is suspended horizontally by two vertical cables at each end. Cable \(A\) can support a maximum tension of 500.0 \(\mathrm{N}\) without breaking, and cable \(B\) can support up to 400.0 \(\mathrm{N} .\) You want to place a small weight on this bar. (a) What is the heaviest weight you can put on without breaking either cable, and (b) where should you put this weight?

A uniform strut of mass \(m\) makes an angle \(\theta\) with the horizontal. It is supported by a frictionless pivot located at one- third its length from its lower left end and a horizontal rope at its upper right end. A cable and package of total weight \(w\) hang from its upper right end. (a) Find the vertical and horizontal components \(V\) and \(H\) of the pivot's force on the strut as well as the tension \(T\) in the rope. (b) If the maximum safe tension in the rope is 700 \(\mathrm{N}\) and the mass of the strut is \(30.0 \mathrm{kg},\) find the maximum safe weight of the cable and package when the strut makes an angle of \(55.0^{\circ}\) with the horizontal. (c) For what angle \(\theta\) can no weight be safely suspended from the right end of the strut?

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