Chapter 11.3, Problem 42PPS

(a)

To determine

To Find: Themechanical advantage of lever.

Answer to Problem 42PPS

  $4$

Explanation of Solution

Given: The mechanical advantage of a lever is $\frac{L_A}{L_R}$ where $L_A$ is the length of the effort arm and $L_R$ is the length of resistance.

  

From the figure given

  $\begin{array}{l}{L}_A=8ft\\ L_R=2ft\end{array}$

Then

  $\frac{L_A}{L_R}=\frac{8}{2}=4$

Hence, the mechanical advantage of the lever is $4$

(b)

To determine

To Find: If the Egyptian worker can apply a force of $180$ pounds, what is the greatest weight he can lift with the lever.

Answer to Problem 42PPS

  $720pounds$

Explanation of Solution

Given: The force placed on the rock is the product of the mechanical advantage and the force applied to the end of the lever.

  

Mechanical advantage of the lever is $4$

And

Egyptian worker can apply a force of $180$ pounds

Then

Greatest weight = $180\times 4=720pounds$

Hence, the greatest weight is $720pounds$

(c)

To determine

To Find: The force.

Answer to Problem 42PPS

  $152.9pounds$

Explanation of Solution

Given: To lift a $535$ pound rock using a $7$ foot lever with the fulcrum $2$ feet from the rock.

  

  $\begin{array}{l}{L}_A=7ft\\ L_R=2ft\end{array}$

Then

  $\frac{L_A}{L_R}=\frac{7}{2}=3.5$

Then

Force= $\frac{weightofrock}{mechanicaladvantage}$

  $=\frac{535}{3.5}=152.9pounds$

Hence, $152.9pounds$