Question

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

• A. F
• B. mg
• C. F+mg
• D. $\sqrt{F^2+m^2{g}^2}$

Answer 1

The correct option is D $\sqrt{F^2+m^2{g}^2}$

Given that,

Mass $=m$

According to free body diagram,

The tension exerted by the string on the block

The vertical components is

$Tsin\theta =mg$ ....(I)

The horizontal component is

$Tcos\theta =F$ .....(II)

From equation (I) and (II)

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

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

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

$si{n}^2\theta +co{s}^2\theta =(\frac{mg}{T}{)}^2+(\frac{F}{T}{)}^2$

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

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