Laws Of Motion Question 149
Question: A block of mass 1 kg is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.5. The magnitude of the force acting upwards at an angle of $ 60{}^\circ $ from the horizontal that will just start the block moving is
Options:
A) $ 5N $
B) $ \frac{20}{2+\sqrt{3}}N $
C) $ \frac{20}{2-\sqrt{3}}N $
D) $ 10N $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ R+F\sin 60{}^\circ =mg $
$ R=mg-\frac{\sqrt{3}F}{2} $ If block just starts moving $ F\cos 60{}^\circ =f=\mu R $ Or $ F+\frac{\sqrt{3}F}{2}=10 $ or $ F=\frac{20}{2+\sqrt{3}} $
BETA
