Question

a lever is being used to lift a 1,000 - newton object. if the lever has an efficiency of 90% and the input force is 200 newtons, what is the minimum ratio of the input displacement to the output displacement \\( \frac{\delta x_{\text{in}}}{\delta x_{\text{out}}} \\)? (1 point)
4.5
0.2
5.6
5

Explanation:

Step1: Recall the formula for efficiency

The efficiency formula for a machine is $\eta=\frac{W_{out}}{W_{in}}=\frac{F_{out}\Delta x_{out}}{F_{in}\Delta x_{in}}$.

Step2: Rearrange the formula to solve for $\frac{\Delta x_{in}}{\Delta x_{out}}$

From $\eta=\frac{F_{out}\Delta x_{out}}{F_{in}\Delta x_{in}}$, we can rewrite it as $\frac{\Delta x_{in}}{\Delta x_{out}}=\frac{F_{out}}{\eta F_{in}}$.

Step3: Substitute the given values

Given $F_{out} = 1000\ N$, $\eta=0.9$, and $F_{in}=200\ N$.
Substitute into the formula: $\frac{\Delta x_{in}}{\Delta x_{out}}=\frac{1000}{0.9\times200}=\frac{1000}{180}\approx5.6$.

Answer:

5.6