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Chapter 8: Problem 28

A weight of \(1200 \mathrm{N}\) rests on a lever at a point \(0.50 \mathrm{m}\) from a support. On the same side of the support, at a distance of $3.0 \mathrm{m}\( from it, an upward force with magnitude \)F$ is applied. Ignore the weight of the board itself. If the system is in equilibrium, what is \(F ?\)

Short Answer

Expert verified
Answer: The magnitude of the upward force required to keep the lever in equilibrium is 200 N.

Step by step solution

01

Identify the forces acting on the lever

There are two forces acting on the lever: Force 1: The weight of \(1200 \mathrm{N}\) at a distance of \(0.50 \mathrm{m}\) from the support. Force 2: The upward force with magnitude \(F\) at a distance of \(3.0 \mathrm{m}\) from the support.
02

Calculate the torque due to each force

Torque is calculated as the product of force and the perpendicular distance from the axis of rotation, which in this case is the support. Therefore, we will calculate the torque produced by each force around the support. Torque 1: \(\tau_1 = (1200 \mathrm{N})(0.50 \mathrm{m})\) Torque 2: \(\tau_2 = (F)(3.0 \mathrm{m})\)
03

Apply the condition for equilibrium

When the system is in equilibrium, the sum of the torques should be equal to zero. Since the torque produced by the weight (Torque 1) is acting in the clockwise direction and the torque produced by the upward force (Torque 2) is acting in the counterclockwise direction, their effect on the lever will be opposite to each other. To keep the lever in equilibrium, the following condition must hold true: \(\tau_1 = \tau_2\) We can substitute the torques calculated in step 2: \((1200 \mathrm{N})(0.50 \mathrm{m}) = (F)(3.0 \mathrm{m})\)
04

Solve for F

To find the magnitude of the upward force \(F\), we can rearrange the equation above and solve for F: \(F = \frac{(1200 \mathrm{N})(0.50 \mathrm{m})}{3.0 \mathrm{m}}\) Now, we can plug in the values and calculate \(F\): \(F = \frac{600 \mathrm{N\cdot m}}{3.0 \mathrm{m}} = 200 \mathrm{N}\)
05

State your answer

The magnitude of the upward force \(F\) required to keep the lever in equilibrium is \(200 \mathrm{N}\).

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