Three boys, each of mass 45 kg, pull simlutaneously a block on a smooth surface. The mass of block is 20.0 kg.
(a) Find the acceleration of the block.
(b) Find the acceleration of the boy exerting force `F_(1)` . Assume no friction between the boy and the surface .

(Given `F_(1)=90N, F_(2)=114 N, and F_(3)=128sqrt(2)N]`.

Explanation

First we will resolve all the force acting on the block into x and y components.

`Sigma F_(x)=F_(1) cos 30^(@) +F_(3) cos 45^(@)`
`Sigma F_(y)=F_(1) sin 30^(@) + f_(2)-F_(3) sin 45 ^(@)`
Now `a_(x)=(Sigma F_(x))/(m) and a_(y) =(Sigma F_(y))/(m)`
`a_(x)=((90.0)cos 37^(@)+128 sqrt(2) cos 45.0^(@))/(20.0) =10 ms^(-2)`
`a_(y)=((90.0)(sin 37^(@))+114-(128sqrt(2)sin 45.0^(@)))/(20.0) = 2ms^(-2)`
According to Newton's law, the force ecerted on the block by the boy must be equal to the force exerted on the body by the block. Therefore.

`(a_1) =(F'_(1))/(m) =(90.0)/(45)==2 ms^(-2)`.