Step 1: Understanding the Question:
Two forces do the same amount of work over the same distance, but they act at different angles. We need to find the ratio of the magnitudes of these forces.
Step 2: Key Formula or Approach:
The work done (\(W\)) by a constant force (\(F\)) that makes an angle \(\theta\) with the displacement (\(d\)) is given by:
\[ W = Fd \cos\theta \]
We are given that the work done by person A (\(W_A\)) is equal to the work done by person B (\(W_B\)).
Step 3: Detailed Explanation:
Let \(F_A\) be the force applied by person A at an angle \(\theta_A = 45^\circ\).
Let \(F_B\) be the force applied by person B at an angle \(\theta_B = 60^\circ\).
The displacement is \(d\) for both.
Work done by A:
\[ W_A = F_A d \cos(45^\circ) \]
Work done by B:
\[ W_B = F_B d \cos(60^\circ) \]
Given \(W_A = W_B\):
\[ F_A d \cos(45^\circ) = F_B d \cos(60^\circ) \]
The displacement \(d\) cancels out:
\[ F_A \cos(45^\circ) = F_B \cos(60^\circ) \]
We need to find the ratio \(\frac{F_A}{F_B}\).
\[ \frac{F_A}{F_B} = \frac{\cos(60^\circ)}{\cos(45^\circ)} \]
Now, substitute the values of the trigonometric functions:
\[ \cos(60^\circ) = \frac{1}{2} \]
\[ \cos(45^\circ) = \frac{1}{\sqrt{2}} \]
So, the ratio is:
\[ \frac{F_A}{F_B} = \frac{1/2}{1/\sqrt{2}} = \frac{1}{2} \times \frac{\sqrt{2}}{1} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \]
The problem states that this ratio is equal to \(\frac{1}{\sqrt{x}}\).
\[ \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{x}} \]
By comparing the two sides, we can see that:
\[ x = 2 \]
Step 4: Final Answer:
The value of x is 2.