Question

"A window washer pulls herself upward using the bucket-pulley apparatus. The mass of the person plus the bucket is 65 kg.

(a) How hard must she pull downward to raise herself slowly at constant speed?
(b) If she increases this force by 10%, what will her acceleration be?
"

Answer 1

Consider the free body diagrams as shown in the file attached.

Let the acc of system is a.
For the washer T+N-m1g=m1a ............................(1)
For the bucket T-N-m2g=m2a ............................(2)

(A)
a=0 Eqns (1) and (2) become T+N-m1g=0 T-N-m2g=0 Adding and rearranging, T=(m1+m2)/2g T=325N
(B)
We are specified the value of T here as 325+32.5=357.5N eqns (1)and (2) become 357.5+N-m1g=m1a 357.5-N-m2g=m2a Adding, 715-(m1+m2)g=(m1+m2)a

Solve for 'a'

Answer 2

Part A.

If there is a constant speed, therefore the net force is zero.

Forces acting are the:

1. Weight

2. Upward force exerted by the rope which is opposite in direction but the magnitude is to the force which is acting downwards exerted on the rope.

∑Force = pulling force - weight = 0

Weight = mg

Hence,

F (pulling force) = mg

Substituting the values,

F (pulling force) = 65 × 9.81

F (pulling force) = 637.65 Newton

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Part B.

F = ma

10% increase in the force = 1.10 ₓ 637.65

10% increase in the force = 701.415

Therefore,

701.415 = 65 ₓ a

a = 701.45/65

a = 10.8 m/s²