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 anlge of 60o
from the horizontal that will just start
the block moving is

Answer 1

Your answers is given below. ..

given m=1kg, weight = mg =10N

R=normal reaction of table on block

coefficient of static friction be u=0.5

applied force(F) splits into 2components..

Fcos60(Horizontal component)

Fsin60(Vertical component along direction of weight)

R=mg+Fsin60

=10+Fsin60

now equating all forces..

Fcos60=uR

Fcos60=(0.5)[mg+Fsin60]

Fcos60=(0.5)[10+Fsin60]

solving F=74.6N