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Chapter 10: Problem 21

A \(110-\mathrm{N} \cdot \mathrm{m}\) torque is needed to start a revolving door rotating. If a child can push with a maximum force of \(90 \mathrm{N},\) how far from the door's rotation axis must she apply this force?

Short Answer

Expert verified
The child must apply the force 1.22 meters away from the door's rotation axis.

Step by step solution

01

Write down given values

The torque \( \tau = 110 \mathrm{N} \cdot \mathrm{m} \) and the force \( F = 90 \mathrm{N} \) are given. The angle \( \theta \) is \(90^{\circ}\) since the child pushes the door perpendicularly to the radius. Thus, \( \sin(\theta) = 1 \). The value to find is the distance \( r \) from the axis of rotation.
02

Rearrange the formula

The formula for torque is \( \tau = rF \sin(\theta) \). By rearranging the formula to solve for \( r \), we get \( r = \frac{\tau}{F \sin(\theta)} \).
03

Plug in the values and calculate \( r \)

Substitute \( \tau = 110 \mathrm{N} \cdot \mathrm{m} \), \( F = 90 \mathrm{N} \), and \( \sin(\theta) = 1 \) into the equation: \( r = \frac{110 \mathrm{N} \cdot \mathrm{m}}{90 \mathrm{N}} = 1.22 \mathrm{m} \).

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Most popular questions from this chapter

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