A man has fallen into a ditch of width d and two of his friends are slowly pulling him out using a light rope and two fixed pulleys as shown in figure (5−E4). Show that the force (assumed equal for both the friends) exerted by each friend on the road increases as the man moves up. Find the force when the man is at a depth h.
Answers
Answer:
The force when the man is at a depth h is mg/ 4h√ [d²+4h²]
Explanation:
(a) At any depth, let the ropes makes an angle θ with the vertical.
From the free-body diagram,
Fcosθ + Fcosθ − mg = 0
2Fcosθ = mg
⇒F=mg/2cosθ
As the man moves up, θ increases, i.e. cosθ decreases.
Thus, F increases.
(b) Now When the man is at depth h,
cosθ=h/ √d/2²+h²
F=mg/2cosθ
=mg/2 [h/√d/2²+h²
=mg/ 4h√ [d²+4h²]
Answer:
At any depth,
Let the ropes makes an angle with the vertical
Then,
From the free body diagram
Fcosθ+Fcosθ−mg=0
2Fcosθ=mg
F=
2cosθ
mg
So, as the man moves up,θ increases, cosθ decreases
Thus, force increases
Now, when the man is at depth h
Then, the force is
cosθ=
(
2
d
)
2
+h
2
h
F=
2×
(
2
d
)
2
+h
2
h
mg
F=
4h
mg
d
2
+4h
2
Hence, the force is
4h
mg
d
2
+4h
2