Laws Of Motion Question 313

Question: An overweight acrobat, “weighing” in at 115 kg, wants to perform a single hand stand. He tries to cheat by resting one foot against a smooth frictionless vertical wall. The horizontal force there is 130 N. What is the magnitude of the force exerted by the floor on his hand? Answer in N.

Options:

A) 1134

B) 1257

C) 997

D) 1119

Show Answer

Answer:

Correct Answer: A

Solution:

[a] The acrobat has a force acting on his hand that we resolve into two perpendicular components:the vertical one is the reaction to the weight $ (115\times 9.8N=1127N) $ and the horizontal one balances the 130N force from the wall.

These two forces give a resultant force F of $ F=\sqrt{1127^{2}+130^{2}}=1134N $



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