Certainly! Let's solve the problem step-by-step.
### Problem Statement
We are given a two-pulley system with an efficiency of 75%. We need to find the effort required to lift a load of 500 Newtons using this pulley system.
### Known Values
1. Efficiency of the system: [tex]\( 75\% \)[/tex]
2. Load to be lifted: [tex]\( 500 \, \text{N} \)[/tex]
### Understanding Efficiency
Efficiency ([tex]\(\eta\)[/tex]) in a mechanical system is defined as the ratio of the useful output work to the input work. For this problem, the efficiency can be expressed as:
[tex]\[
\eta = \frac{\text{Load (or Output Work)}}{\text{Effort (or Input Work)}} \times 100\%
\][/tex]
### Converting Efficiency to Decimal Form
Since efficiency is given as a percentage, we convert it to a decimal for ease of calculation:
[tex]\[
\eta = 75\% = 0.75
\][/tex]
### Formula for Effort
To find the effort required, we rearrange the efficiency formula to solve for effort ([tex]\(E\)[/tex]):
[tex]\[
\text{Efficiency} = \frac{\text{Load}}{\text{Effort}}
\][/tex]
[tex]\[
0.75 = \frac{500 \, \text{N}}{\text{Effort}}
\][/tex]
We need to isolate "Effort" (E) on one side:
[tex]\[
\text{Effort} = \frac{500 \, \text{N}}{0.75}
\][/tex]
### Calculating the Effort
Now, let's compute the value:
[tex]\[
\text{Effort} = \frac{500 \, \text{N}}{0.75} = 666.67 \, \text{N}
\][/tex]
### Final Answer
The effort required to lift a load of 500 N with an efficiency of 75% is approximately [tex]\(666.67 \, \text{N}\)[/tex].
So, the detailed calculation for finding the effort required to lift a load of 500 N with a two-pulley system, given the system's efficiency is 75%, yields an effort of [tex]\( 666.67 \, \text{N} \)[/tex].
