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 $
BETA
