Question

2. A block of mass M is pulled on a smooth
horizontal table by a string making an
angle 9 with the horizontal as shown in
figure. If the acceleration of the block is a.
find the force applied by the string and by
the table on the block.​

Answer 1

Answer: Let's assume the angle with which string is pulled and horizontal to be x

And the force with which it is pulled to be F

So the horizontal component of this force will be Fcosx

Which is responsible for the acceleration 'a' of the block

So,

Fcosx=Ma

F= Ma/cosx

Now considering the vertical forces acting on block

1.) Mg (downwards)

2.) Fsinx ( vertical component of force applied by string)

So total vertical force acting= Mg-Fsinx

This vertical force will be cancelled by the upward normal force applied by table on the block

N= Mg-Fsinx