To paint the side of a building, painter normally hoists himself up by pulling on the rope A as in figure. The painter and plateform together weigh 200 N. The rope B can withstand 300 N. Then

Options

A. The maximum acceleration that painter can have upwards is `5(m)/(s^2)`.
B. To hoist himself up, rope B must withstand minimum 400 N force.
C. Rope A will have a tension of 100 N when the painter is at rest.
D. The painter must exert a force of 200 N on the rope A to go downwards slowly.

Answer

A, C

Explanation

Maximum force, the rope B can withstand is 300 N, so the maximum tension which can be produce in rope A should be `=(300)/(2)=150N`
(i) For maximum acceleration:
`150+150-200=20a_(max)`
`a_(max)=5(m)/(s^2)`
(ii) Now to hoist up (say with uniform velocity), minimum tension in A should be 100 N (or painter should apply force of 100 N on rope A), then minimum tension in B should be 200N. So B should withstand minimum tensin of 200 N. Same is the case to move down slowly.
(iii) When the painter is at rest the tension in A should be 100 N to balance the weight of paiter and platform. So painter will exert 100 N on rope A.