As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30 ° , with horizontal. For μ s = 0 . 25 , the block will just start to move for th

Question

As shown in the figure a block of mass 10 kg lying on a horizontal surface is pulled by a force F acting at an angle 30 ° , with horizontal. For μ s = 0 . 25 , the block will just start to move for the value of F : [Given g = 10 m · s – 2 ]

TestHub · Accepted Answer

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

$\begin{array}{rcl} N & = & m g - F \sin 30 ° \\ = & m g - \frac{F}{2} \\ = & 100 - \frac{F}{2} \\ = & \frac{200 - F}{2} . . . . . . . . . . . . . . . . . . . . . . . . (1) \end{array}$

From the equilibrium of horizontal component of forces, it can be written that,

$F \cos 30 ° = μ N . . . . . . . . . . . . . . . . . . . . . . (2)$

Substitute the expression for the normal reaction force from equation (1) into equation (2) and solve to calculate the value of the applied force.

$\begin{array}{rcl} \sqrt{3} \frac{F}{2} & = & 0.25 \times (\frac{200 - F}{2}) \\ 4 \sqrt{3} F & = & 200 - F \\ F & = & \frac{200}{4 \sqrt{3} + 1} \\ \approx & 25.2 \end{array}$

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