Answer to Question #124626 in Other for Lanbonbik Jerry

Question

A particle of mass 10 kg is attached to one end of a light inextensible string whose other end
is fixed. The particle is pulled aside by a horizontal force until the string is at 60° to the
vertical. Find the magnitude of the horizontal force and the tension in the string.

Answer 1

According to second Newton's Law

$$\vec{F}+m\vec{g}+\vec{T}=0$$

$$Oy: T\cos\theta-mg=0$$

$$Ox: F-T\sin\theta=0$$

$$T={mg\over \cos \theta}$$

$$F=mg\tan \theta$$

$$T={10kg\cdot9.81 m/s^2\over \cos 60 \degree}=196.2\ N$$

$$F=10kg\cdot9.81m/s^2\tan60\degree=98.1\sqrt{3} \ N\approx169.9\ N$$