Question

A wooden table of mass 100 kg is lying
on a floor. A man pushes the table with
a force of 100 N to the right and another
man tends to stop the table with a force
of 200 N. If the friction between the
floor and table is 80 N, then​

Answer 1

a) The tension in wire T attached to the hook can be determined as,

Tsin(θ)w

T=sin(θ)w

The force due to static friction on block A is equal to the horizontal component of the

tension.

The expression for frictional force is,

Ff=Tcos(θ)

Substitute T=sin(θ)w in the frictional force expression

Ff=sin(θ)wcos(θ)

=wcot(θ)

Substitute 12 N for w and 45∘ for θ.

Ff=(12N)cot(45∘)

=12N

Therefore friction force exerted on block A is 12N.

(b)

From the system to remain in equilibrium force of friction should be eqy=ual to the weight w of the block.

The frictional forceis,

Ff=mAgμ2

Here, mA is mass of A, g is acceleration due to gravity and μs is coefficient

of static friction.

Substitute 60 N for mg and 0.25 for μs.

Ff=(60N)(0.25)

=15N

The force of friction should be equal to the weight w of the block so the weight of the 

block is also 15 N.

Therefore, maximum weight w for the system to remain in equilibrium is 15 N.