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A block of mass 1kg is at rest on a hori...

A block of mass 1kg is at rest on a horizontal table. The coeficient of static friction between the block and the table is `0.5` . The magnitude of the force acting upward at an angle of `60^(@)` from the horizontal that will just start the block moving is.

A

5 N

B

5.36 N

C

74.6 N

D

10 N

Text Solution

Verified by Experts

The correct Answer is:
B


From the figure, R+F `sin60^(@)=mg`
`:." "R=mg-Fsin60^(@)=mg-(sqrt3)/(2)F`
When the block just starts moving. The limiting force of friction, `f=muR`
and f=applied force=`Fcos60^(@)`
`:." "Fcos60^(@)=mu[mg-(sqrt3)/(2)F]`
`:." " F/2=1/2[1xx10-(sqrt3)/(2)F]`
`:." "F=10-(sqrt3)/(2)F`
`:." "F+(sqrt3)/(2)F=10" ":." "2F+sqrt3F=20`
`:." "F=(20)/(2+sqrt3)=(20)/(2+1.732)=(20)/(3.732)`
`:." "F=5.36 N`
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