Question

In the following figure the object of mass m is held at rest by the horizontal force as shown. The force exerted by the string on the block is

Answer 1

Answer:

The force exerted by the string on the block is $T=\sqrt{(mg)^2+(F)^2}$

Explanation:

Given that,

mass = m

According to FBD,

The tension exerted by the string on the block

The vertical components is

$T\sin\theta= mg$....(I)

The horizontal component is

$T\cos\theta=F$.....(II)

From equation (I) and (II)

$\sin\theta=\dfrac{mg}{T}$....(III)

$\cos\theta=\dfrac{F}{T}$.....(IV)

On squaring and adding equation (III) and (IV)

$\sin^{2}\theta+\cos^2\theta=(\dfrac{mg}{T})^2+(\dfrac{F}{T})^2$

$T=\sqrt{(mg)^2+(F)^2}$

Hence, The force exerted by the string on the block is $T=\sqrt{(mg)^2+(F)^2}$

Answer 2

Explanation:

This is the answer in short