Question

A body of mass 10 kg is placed on rough surface, pulled by a force F having an angle 30
degree above the horizontal . If the angle of friction is also 30 degree, then the minimum magnitude of force F required to move the body is equal to (g= 10ms^-2) ?

Answer 1

Dear student,
Downward force acting on the object $=mg\sin \theta +\mu mg\cos \theta$
here, $m=10kg$, $\mu =0.5773502692$, $\theta ={30}^0$
So the downward force is
$$\begin{gathered} =10\times 10\times \sin \left({30}^0\right)+0.5773502692\times 10\times 10\times \cos \left({30}^0\right) \\ =10\times 10 \\ =100N \end{gathered}$$

Hence the minimum force needed to move the block is 100N.
Regards