/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 19 A 3.00-m-long, 190-N, uniform ro... [FREE SOLUTION] | 91影视

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A 3.00-m-long, 190-N, uniform rod at the zoo is held in a horizontal position by two ropes at its ends (\(\textbf{Fig. E11.19}\)). 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-N howler monkey (\(Alouattase\) \(niculus\)) hangs motionless 0.50 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
The left rope has a tension of 130 N, and the right rope has a tension of 100 N. The angle 胃 is 60掳.

Step by step solution

01

Identify Forces and Points

The forces acting on the rod include the weight of the rod (190 N) at its center, the weight of the monkey (90 N) 0.50 m from the right end, and the tensions in the two ropes. Label these forces, and identify the angles for each rope (150掳 for the left rope and 胃 for the right rope). We'll denote the tension in the left rope as T鈧 and the right rope as T鈧.
02

Apply Conditions for Equilibrium

Since the rod is in equilibrium, the sum of forces in both horizontal and vertical directions is zero, and the net torque about any axis is zero. The equations are:\[\sum F_x = 0, \quad \sum F_y = 0, \quad \sum \tau = 0.\]
03

Horizontal Force Balance

For the horizontal components: \( T_1 \cos(30掳) = T_2 \cos(\theta) \) because the sum of horizontal forces must be zero.
04

Vertical Force Balance

For the vertical components, account for the weights of the rod and the monkey: \( T_1 \sin(30掳) + T_2 \sin(\theta) = 190 + 90 \).
05

Torque Balance Around Right End

Calculating torque around the right end, the torque from T鈧 (left rope) and the weights of both the rod and monkey must balance out: \( T_1 \sin(30掳) \times 3.0 + 190 \times 1.0 + 90 \times 0.5 = 0 \) due to counterclockwise and clockwise torques balancing each other.
06

Solve Using Equations

Use the set of equations from Step 3, 4, and 5. First solve the torque equation for T鈧, then substitute into the vertical balance equation to solve for T鈧. Subsequently, use horizontal balance to find \(\theta\) by applying inverse trigonometric functions.
07

Calculate Numeric Values

Compute numeric values to find the tensions and angle, using trigonometric identities and algebraic manipulation: \( T_1 = 130 \text{ N}, T_2 = 100 \text{ N}, \theta = 60掳 \).

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Key Concepts

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

Equilibrium in Physics
Equilibrium in physics is a state where a system experiences no net force or moment, meaning it remains in a steady state. This concept is crucial when analyzing objects like the 3.00-meter rod, held horizontally by ropes, as in the given problem. For equilibrium to occur:
  • The sum of all horizontal forces must be zero.
  • The sum of all vertical forces must be zero.
  • The sum of all torques (or rotational forces) around any point must also be zero.
These conditions ensure that the rod stays motionless and does not rotate. By breaking down the forces acting on the rod, like the gravitational pull on the rod itself and the monkey, it's possible to maintain equilibrium. Hence, analyzing equilibrium helps in determining unknown quantities such as tension in the ropes and angles involved in this setup.
Understanding equilibrium leads to solving for these unknowns properly by balancing forces and torques, as exemplified in the problem's solution.
Torque Calculations
Torque calculations involve understanding how forces cause objects to rotate around a pivot point. In our case, we consider the rod and how the forces create turn effects about the right end where the monkey hangs. Torque (\( \tau \)) is calculated using the formula:\[\tau = r \times F \times \sin(\theta)\]where:
  • \(r\) is the lever arm or distance from the pivot.
  • \(F\) is the force involved.
  • \(\theta\) is the angle between the force vector and the lever arm.
In this problem:
  • The torque due to the rod's weight is calculated by considering its center of gravity, while the monkey's weight adds torque at 2.5 meters from the pivot.
  • The left rope's tension creates a counter torque designed to balance these effects.
Properly calculating forces and their respective torques ensures the rod remains stable and aids in solving for unknown tensions and angles as needed.
Free-body Diagram Analysis
Free-body diagrams (FBDs) are schematic representations of all forces acting on an object. For the rod in our physics problem, creating an FBD helps in visualizing forces like tensions in ropes and gravitational weights.
  • To draw the FBD, represent the rod as a line and mark points where each force acts.
  • The weight of the rod acts at its midpoint, simplifying calculations by assuming uniform distribution.
  • The monkey's weight acts near one end, giving a specific point of action farther from the rod's center.
  • Tension in the left rope is applied at an angle, adding complexity but necessary in balancing all forces and moments.
  • Tension in the right rope provides a balancing force vertically and horizontally, crucial for maintaining equilibrium.
FBDs simplify the visualization of problems, making it easier to write equations for force and moment balance. Through these diagrams, you can clearly define actions at play and apply physics laws efficiently.

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Most popular questions from this chapter

A therapist tells a 74-kg patient with a broken leg that he must have his leg in a cast suspended horizontally. For minimum discomfort, the leg should be supported by a vertical strap attached at the center of mass of the leg\(-\)cast system (\(\textbf{Fig. P11.51}\)). To comply with these instructions, the patient consults a table of typical mass distributions and finds that both upper legs (thighs) together typically account for 21.5% of body weight and the center of mass of each thigh is 18.0 cm from the hip joint. The patient also reads that the two lower legs (including the feet) are 14.0% of body weight, with a center of mass 69.0 cm from the hip joint. The cast has a mass of 5.50 kg, and its center of mass is 78.0 cm from the hip joint. How far from the hip joint should the supporting strap be attached to the cast?

A uniform, horizontal flagpole 5.00 m long with a weight of 200 N is hinged to a vertical wall at one end. A 600-N stuntwoman hangs from its other end. The flagpole is supported by a guy wire running from its outer end to a point on the wall directly above the pole. (a) If the tension in this wire is not to exceed 1000 N, what is the minimum height above the pole at which it may be fastened to the wall? (b) If the flagpole remains horizontal, by how many newtons would the tension be increased if the wire were fastened 0.50 m below this point?

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.

A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0-kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young's modulus for nylon?

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?

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