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### College Physics (Urone)

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 $$784\;{\rm{N}}$$.

See the step by step solution

## Step 1: Given Data

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 $$3\;{\rm{m}}$$.

## Step 2: Calculation of the force

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}