Question

A block weighing 10kg is at rest on a horizontal
table. The coefficient of static friction between
the block and the table is 0.5. If a force acts
down at 60° from the horizontal, how large
can it be without causing the block to move? (
g= 10 ms?)
​

Answer 1

Considering horizontal forces being balanced just before the motion of the block,

Fcos 60=ms*N

F/2=0.5*N

N=weight of block+vertical component of force applied

=100+F sin60

=100+√3F/2

therefore,

F/2=(1/2)*(100+√3F/2)

2F=200+√3F

Therefore,

F=200/(2-√3)

Answer 2

Answer:

don't know sorry

Explanation:

btw thnx for extra points