A mass of 10kg is suspended by a rope of length 4m , from the...

Let the tension in the string is $T$.

Applying the condition of equilibrium in the vertical direction,

Tcos\left(45 °\right) = 100$</p><p>$⇒ \dfrac{T}{\sqrt{2}} = 100 . . . \left(1\right)$</p><p>In the horizontal direction,</p><p>Tsin\left(45 °\right) = F$

$⇒ \dfrac{T}{\sqrt{2}} = F$

Put the value of $T$ from equation \left(1\right)$,</p><p>$⇒ F = 100N$</p><p>So, the horizontal force applied on the rope will be$100N$.