Question 86 of 210

As shown in the figure a block of mass $10kg$ lying on a horizontal surface is pulled by a force F acting at an angle $30^{\circ}$, with horizontal. For ${\mu }_{s}=0.25$, the block will just start to move for the value of $F$: [Given g $=10m\cdot {s}^{–2}$]

A$33.3N$B$25.2N$C$20N$D$35.7N$

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Answer: (B) $25.2N$

The free body diagram for the given scenario is shown below-
From the equilibrium of the vertical component of forces, it can be written that
$\begin{matrix}N & = & mg-F\sin 30^{\circ} \\ & = & mg-\dfrac{F}{2} \\ & = & 100-\dfrac{F}{2} \\ & = & \dfrac{200-F}{2}........................(1)\end{matrix}$
From the equilibrium of horizontal component of forces, it can be written that,
$F\cos 30^{\circ}=\mu N......................(2)$
Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.
$\begin{matrix}\sqrt{3}\dfrac{F}{2} & = & 0.25\times \left(\dfrac{200-F}{2}\right) \\ 4\sqrt{3}F & = & 200-F \\ F & = & \dfrac{200}{4\sqrt{3}+1} \\ & \approx & 25.2\end{matrix}$