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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 playground merry-go-round of radius \(2.00 \mathrm{~m}\) has a moment of inertia \(I=275 \mathrm{~kg} \cdot \mathrm{m}^{2}\) and is rotating about a frictionless vertical axle. As a child of mass \(25.0 \mathrm{~kg}\) stands at a distance of \(1.00 \mathrm{~m}\) from the axle, the system (merrygo-round and child) rotates at the rate of \(14.0 \mathrm{rev} / \mathrm{min}\). The child then proceeds to walk toward the edge of the merry-go-round. What is the angular speed of the system when the child reaches the edge?

An Atwood's machine consists of blocks of masses \(m_{1}=10.0 \mathrm{~kg}\) and \(m_{2}=20.0 \mathrm{~kg}\) attached by a cord running over a pulley as in Figure \(\mathrm{P} 8.40 .\) The pulley is a solid cylinder with mass \(M=8.00 \mathrm{~kg}\) and radius \(r=0.200 \mathrm{~m}\) The block of mass \(m_{2}\) is allowed to drop, and the cord turns the pulley without slipping. (a) Why must the tension \(T_{2}\) be greater than the tension \(T_{1} ?\) (b) What is the acceleration of the system, assuming the pulley axis is frictionless? (c) Find the tensions \(T_{1}\) and \(T_{2}\).

A \(12.0\) -kg object is attached to a cord that is wrapped around a wheel of radius \(r=10.0 \mathrm{~cm}\) (Fig. P8.70). The acceleration of the object down the frictionless incline is measured to be \(2.00 \mathrm{~m} / \mathrm{s}^{2}\). Assuming the axle of the wheel to be frictionless, determine (a) the tension in the rope, (b) the moment of inertia of the wheel, and (c) the angular speed of the wheel \(2.00 \mathrm{~s}\) after it begins rotating, starting from rest.

A bicycle wheel has a diamcter of \(64.0 \mathrm{~cm}\) and a mass of \(1.80 \mathrm{~kg}\). Assume that the wheel is a hoop with all the mass concentrated on the outside radius. The bicycle is placed on a stationary stand, and a resistive force of \(120 \mathrm{~N}\) is applied tangent to the rim of the tire. (a) What force must be applied by a chain passing over a \(9.00-\mathrm{cm}-\) diameter sprocket in order to give the wheel an acceleration of \(4.50 \mathrm{rad} / \mathrm{s}^{2} ?\) (b) What force is required if you shift to a \(5.60-\mathrm{cm}\) -diameter sprocket?

The large quadriceps muscle in the upper leg terminates at its lower end in a tendon attached to the upper end of the tibia (Fig. \(\mathrm{P} 8.29 \mathrm{a}\) ). The forces on the lower leg when the leg is extended are modeled as in Figure P8.29b, where \(\vec{T}\) is the force of tension in the tendon, \(\vec{w}\) is the force of gravity acting on the lower leg, and \(\overrightarrow{\mathbf{F}}\) is the force of gravity acting on the foot. Find \(\vec{T}\) when the tendon is at an angle of \(25.0^{\circ}\) with the tibia, assuming that \(w=30.0 \mathrm{~N}\), \(F=12.5 \mathrm{~N}\), and the leg is extended at an angle \(\theta\) of \(40.0^{\circ}\) with the vertical. Assume that the center of gravity of the lower leg is at its center and that the tendon attaches to the lower leg at a point one-fifth of the way down the leg
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