Question

An accident victim with a broken leg is being placed in traction. The patient wears a special boot with a pulley attached to the sole. The foot and boot together have a mass of 4.0 kg, and the doctor has decided to hang a m = 6.3 kg mass from the rope. The boot is held suspended by the ropes and does not touch the bed.

A)

Determine the amount of tension in the rope by using Newton’s laws to analyze the hanging mass. Assume that ϕ = 11 ∘.
Hint: If the pulleys are frictionless, which we will assume, the tension in the rope is constant from one end to the other.

Express your answer to two significant figures and include the appropriate units.

B)

The net traction force needs to pull straight out on the leg. What is the proper angle θ for the upper rope?

Express your answer to two significant figures and include the appropriate units.

C)

What is the net traction force pulling on the leg?

Express your answer to two significant figures and include the appropriate units.

Answer 1

(A) The tension in the rope:

T = m*g

= 6.3*9.8

= 61.74 N

= 62 N

(B) For the leg to be pulled straight out, the vertical components of forces on the leg must sum to zero.

The vertical component from the lower part of rope is -T*sin11º

The vertical component from the upper part of the rope is T*sinθ;

The vertical force from the weight of the foot and boot is -4.0*9.8 = -39.2 N

Then, summing:

-T*sin11º + T*sinθ - 39.2 = 0

sinθ = (39.2 + 61.74*sin111º)/61.74 = 0.826

θ = 55.66º

= 56 º

(C) The net traction force is the sum of the horizontal components:

T*cosθ + T*cos11º

= 95 N

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