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Chapter 8: Problem 20

A window washer is standing on a scaffold supported by a vertical rope at each end. The scaffold weighs \(200 \mathrm{~N}\) and is \(3.00 \mathrm{~m}\) long. What is the tension in each rope when the \(700-\mathrm{N}\) worker stands \(1.00 \mathrm{~m}\) from one end?

Short Answer

Expert verified
Doing the math yields a tension of 466.7 N in the rope near the worker and a tension of 433.3 N in the far rope. This conclusion is reasonable, as the worker standing closer to one rope puts more weight on that rope.

Step by step solution

01

Calculate Torque

Torques are calculated by the formula \(\tau = rF\sin(\theta)\), where \(r\) is the distance from the pivot point to where the force is applied, \(F\) is the force, and \(\theta\) is the angle between the force vector and the lever arm vector. In this case, the angle between the force and the lever arm is 90 degrees, so the \(\sin(\theta)\) is 1. Thus, the torque can be simplified to \(\tau=rF\). Because the system is in equilibrium, the sum of the torques should be zero. Set the pivot point at the location of the worker. Then the torques are \(r_1*F_1 = r_2*F_2\), where \(r_1=1.00m\) is the distance from the left rope to the worker, \(F_1\) is the tension in the left rope, \(r_2=2.00m\) is the distance from pivot point to the right rope and \(F_2\) is the force exerted by the right rope.
02

Calculate Force

Total force on the scaffold is zero, because the scaffold is in equilibrium. If \(F_1\) and \(F_2\) are tensions in the left and right ropes respectively, the total force on the scaffold will be \(F_1 + F_2 - F_{scaffold} - F_{worker} = 0\), where \(F_{scaffold}=200N\) and \(F_{worker}=700N\).
03

Solve the Equations

Now, there are two equations and two unknowns. Solve this simultaneous equation to get the tension in each rope.
04

Calculate the Values

Solve the equation \(r_1*F_1 = r_2*F_2\) to get \(F_1 = (r_2 / r_1) * F_2\). Substitute \(F_1\) in the equation \(F_1 + F_2 - F_{scaffold} - F_{worker} = 0\) and solve for \(F_2\). Then substitute \(F_2\) back into the equation \(F_1 = (r_2 / r_1) * F_2\) to find \(F_1\).

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Key Concepts

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

Equilibrium in Physics
In physics, when we discuss equilibrium, we're referring to the state in which all forces acting upon an object are balanced, resulting in no acceleration of that object. To understand this concept, imagine balancing a book on your head; for the book to stay steady, the upward force from your head must equal the gravitational force pulling the book downward. If these forces are unbalanced, the book will fall. Similarly, the window washer on the scaffold is in a state of equilibrium because the forces of tension from the ropes and the gravitational forces (from the scaffold鈥檚 and the worker's weight) exactly cancel each other out.
Torque Calculation
Torque is a measure of the turning force on an object such as a bolt or a beam. It is calculated as the product of the force applied and the distance from the pivot point at which it's applied, considerng direction as a factor. The proper equation is \(\tau = rF\sin(\theta)\), with \(\tau\) representing torque, \(r\) the distance to the pivot, \(F\) the applied force, and \(\theta\) the angle between the force direction and the lever arm. When the angle is 90 degrees, as it commonly is in these problems, the equation simplifies to \(\tau = rF\) because \(\sin(90^\circ) = 1\). In a scenario of static equilibrium, the sum of all torques around any pivot point is zero, which is a pivotal concept when solving for unknown forces in physics problems.
Tension in Ropes
Tension refers to the force conducted along the length of a flexible connector, such as a rope or cable, when it is pulled tight by forces acting from opposite ends. It is experienced as a force that attempts to restore the rope to its unstretched or 'relaxed' length. Tension is always directed along the rope and pulls equally on the objects on either end of the rope. In the problem of the window washer on the scaffold, the tension in the ropes supporting the scaffold is what keeps the system in static equilibrium, balancing the gravitational forces.
Static Equilibrium
Static equilibrium is a particular state of equilibrium where an object is at rest and remains at rest, meaning it has zero velocity and zero acceleration. For an object to be in static equilibrium, not only do the forces need to balance out (as is the case for equilibrium in general), but the torques around any pivot point must also sum to zero. This ensures that the object does not turn or rotate. An understanding of static equilibrium is crucial when analyzing problems like the window washer scenario, where we need to ensure that the scaffold does not tip or accelerate, hence maintaining its state of rest.

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Most popular questions from this chapter

A \(60.0-\mathrm{kg}\) woman stands at the rim of a horizontal turntable having a moment of inertia of \(500 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(2.00 \mathrm{~m}\). The turntable is initially at rest and is free to rotate about a frictionless, vertical axle through its center. The woman then starts walking around the rim clockwise (as viewed from above the system) at a constant speed of \(1.50 \mathrm{~m} / \mathrm{s}\) relative to Earth. (a) In what direction and with what angular speed does the turntable rotate? (b) How much work does the woman do to set herself and the turntable into motion?

A \(4.00-\mathrm{kg}\) mass is connected by a light cord to a \(3.00-\mathrm{kg}\) mass on a smooth surface (Fig. \(\mathrm{P} 8.87\) ). The pulley rotates about a frictionless axle and has a moment of inertia of \(0.500 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and a radius of \(0.300 \mathrm{~m}\). Assuming that the cord does not slip on the pulley, find (a) the acceleration of the two masses and (b) the tensions \(T_{1}\) and \(T_{2}\).

]A \(150-\mathrm{kg}\) merry-go-round in the shape of a uniform, solid, horizontal disk of radius \(1.50 \mathrm{~m}\) is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force must be exerted on the rope to bring the merry-go-round from rest to an angular speed of \(0.500\) rev/s in \(2.00 \mathrm{~s}\) ?

A \(40.0\) -kg child stands at one end of a \(70.0\) -kg boat that is \(4.00 \mathrm{~m}\) long (Fig. \(\mathrm{P} 8.69\) ). The boat is initially \(3.00 \mathrm{~m}\) from the pier. The child notices a turtle on a rock beyond the far end of the boat and proceeds to walk to that end to catch the turtle. (a) Neglecting friction between the boat and water, describe the motion of the system (child plus boat). (b) Where will the child be relative to the pier when he reaches the far end of the boat? (c) Will he catch the turtle? (Assume that he can reach out \(1.00 \mathrm{~m}\) from the end of the boat.)

A merry-go-round rotates at the rate of \(0.20 \mathrm{rev} / \mathrm{s}\) with an \(80-\mathrm{kg}\) man standing at a point \(2.0 \mathrm{~m}\) from the axis of rotation. (a) What is the new angular speed when the man walks to a point \(1.0 \mathrm{~m}\) from the center? Assume that the merry-go-round is a solid \(25-\mathrm{kg}\) cylinder of radius \(2.0 \mathrm{~m} .\) (b) Calculate the change in kinetic energy due to the man's movement. How do you account for this change in kinetic energy?
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