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Chapter 8: Problem 27

A rod is being used as a lever as shown. The fulcrum is \(1.2 \mathrm{m}\) from the load and \(2.4 \mathrm{m}\) from the applied force. If the load has a mass of \(20.0 \mathrm{kg}\). what force must be applied to lift the load?

Short Answer

Expert verified
Solution: To lift the load, a force of 98.1 N must be applied to the rod.

Step by step solution

01

Convert mass to weight

First, we need to convert the mass of the load (\(20.0 \mathrm{kg}\)) to weight, which is the force acting on the load due to gravity, by multiplying the mass by the acceleration due to gravity (\(g\approx 9.81 \mathrm{m/s^2}\)). Weight of the load \(= m \times g = 20.0 \mathrm{kg} \times 9.81 \mathrm{m/s^2} = 196.2 \mathrm{N}\)
02

Calculate the torque due to the load

We can now calculate the torque due to the load as the product of the weight of the load and the distance from the fulcrum: \(\tau_\text{load} = \text{Weight of the load} \times \text{Distance from the fulcrum} = 196.2 \mathrm{N} \times 1.2 \mathrm{m} = 235.44 \mathrm{Nm}\)
03

Find the applied force

In order to lift the load, the applied force must generate a torque equal in magnitude to that of the load but with an opposite direction. Since we know the distance of the applied force from the fulcrum (\(2.4 \mathrm{m}\)), we can calculate the required force: \(\tau_\text{applied} = \tau_\text{load} = F_\text{applied} \times \text{Distance from the fulcrum}\) \(F_\text{applied} = \frac{\tau_\text{load}}{\text{Distance from the fulcrum}} = \frac{235.44 \mathrm{Nm}}{2.4 \mathrm{m}} = 98.1 \mathrm{N}\) So, to lift the load, a force of \(98.1 \mathrm{N}\) must be applied to the rod.

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Most popular questions from this chapter

A hoop of \(2.00-\mathrm{m}\) circumference is rolling down an inclined plane of length \(10.0 \mathrm{m}\) in a time of \(10.0 \mathrm{s} .\) It started out from rest. (a) What is its angular velocity when it arrives at the bottom? (b) If the mass of the hoop, concentrated at the rim, is \(1.50 \mathrm{kg},\) what is the angular momentum of the hoop when it reaches the bottom of the incline? (c) What force(s) supplied the net torque to change the hoop's angular momentum? Explain. [Hint: Use a rotation axis through the hoop's center. \(]\) (d) What is the magnitude of this force?
The radius of a wheel is \(0.500 \mathrm{m} .\) A rope is wound around the outer rim of the wheel. The rope is pulled with a force of magnitude $5.00 \mathrm{N},$ unwinding the rope and making the wheel spin CCW about its central axis. Ignore the mass of the rope. (a) How much rope unwinds while the wheel makes 1.00 revolution? (b) How much work is done by the rope on the wheel during this time? (c) What is the torque on the wheel due to the rope? (d) What is the angular displacement \(\Delta \theta\), in radians, of the wheel during 1.00 revolution? (e) Show that the numerical value of the work done is equal to the product \(\tau \Delta \theta\)
A stone used to grind wheat into flour is turned through 12 revolutions by a constant force of \(20.0 \mathrm{N}\) applied to the rim of a 10.0 -cm-radius shaft connected to the wheel. How much work is done on the stone during the 12 revolutions?
A mechanic turns a wrench using a force of \(25 \mathrm{N}\) at a distance of \(16 \mathrm{cm}\) from the rotation axis. The force is perpendicular to the wrench handle. What magnitude torque does she apply to the wrench?
A flywheel of mass 182 kg has an effective radius of \(0.62 \mathrm{m}\) (assume the mass is concentrated along a circumference located at the effective radius of the flywheel). (a) How much work is done to bring this wheel from rest to a speed of 120 rpm in a time interval of 30.0 s? (b) What is the applied torque on the flywheel (assumed constant)?
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