Astone is rubbed across the horizontal ceiling of a cave passageway (Fig. 6-48). If the coefficient of kinetic friction is 0.65and the force applied to the stone is angled at $\mathit{\theta }\mathbf{=}\mathbf{70}\mathbf{.}{\mathbf{0}}^{\mathbf{°}}$, what must the magnitude of the force be for the stone to move at constant velocity?

• Mass of stone, m = 5.00kg.
• Coefficient of friction, ${\mu }_k=0.65$.
• The angle of force,$\theta =70.0^{°}$

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.

Draw the free body diagram of the stone. Then, applying Newton’s second law, the given problem can be solved.

The free-body diagram for the stone is shown below, with $\stackrel{\rightharpoonup }{F}$being the force applied to the stone, ${\stackrel{\rightharpoonup }{F}}_N$the normal downward force of the ceiling on the stone, $m\stackrel{\rightharpoonup }{g}$the force of gravity, and $\stackrel{\rightharpoonup }{f}$ the force of friction. We take the +x direction to be horizontal to the right and the +y direction to be up.

The equations for the x and the y components of the force according to Newton's second law are:

$\begin{array}{rcl}{F}_x& =& F\cos \theta -f\\ & =& ma\\ F_y& =& F\sin \theta -F_N-mg\\ & =& 0\end{array}$

Friction can be written as,

$$\begin{gathered} f={\mu }_k{F}_N \\ f={\mu }_k\left(F\sin \theta -mg\right) \end{gathered}$$

This expression is substituted for f in the first equation to obtain as:

$F\cos \theta -{\mu }_k\left(F\sin \theta -mg\right)=ma.$

For a = 0, the force is,

$F=\frac{-{\mu }_{k}mg}{\cos \theta -{\mu }_k\sin \theta }$

$$\begin{gathered} F=\frac{-0.65\times 5.0kg\times 9.8m/s^2}{\cos {70}^{°}-0.65\sin {70}^{°}} \\ F=118N \end{gathered}$$

Thus, the magnitude of the force must be 118 N for the stone to move at constant velocity.