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Q17PE

Expert-verifiedFound in: Page 317

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**To get up on the roof, a person (mass \(70.0\;{\rm{kg}}\)) places a \(6.00\;{\rm{m}}\)aluminum ladder (mass \(10.0\;{\rm{kg}}\)) against the house on a concrete pad with the base of the ladder \(2.00\;{\rm{m}}\)from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is \(2\;{\rm{m}}\)from the bottom. The person is standing \(3\;{\rm{m}}\) from the bottom. What are the magnitudes of the forces on the ladder at the top and bottom?**

The magnitude of the force at the top is \(133\;{\rm{N}}\) and at the bottom is \(784\;{\rm{N}}\).

Given:

The mass of the person is \({m_1} = 70.0\;{\rm{kg}}\).

The mass of the ladder is \({m_2} = 10.0\;{\rm{kg}}\).

The length of the ladder is \(l = 6\;{\rm{m}}\).

The distance between the ladder and the concrete pad is \(2.0\;{\rm{m}}\).

The distance between the bottom of the ladder and the person is \(3\;{\rm{m}}\).

The diagram of the ladder can be shown as below,

Applying the condition of equilibrium under forces we get,

\(\begin{align}n - {m_1}g - mg &= 0\\n = {m_1}g + mg\\n &= 70.0 \times 9.8 + 10.0 \times 9.8\\n &= 784\;{\rm{N}}\end{align}\)

Applying the condition of equilibrium under torque, we get,

\(\begin{align}mg{l_1}\cos \theta - mg{l_2}\cos \theta + f\sqrt {{l^2} - {d^2}} - nl\cos \theta &= 0\\f &= \frac{{nl\cos \theta - mg{l_1}\cos \theta + mg{l_2}\cos \theta }}{{\sqrt {{l^2} - {d^2}} }}\\f &= \frac{{784 \times 2 - 10 \times 9.8 \times 4 \times \frac{2}{6} - 70 \times 9.8 \times 3 \times \frac{2}{6}}}{{\sqrt {{6^2} - {2^2}} }}\\ &= 133\;{\rm{N}}\end{align}\)

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