Answer to Question #137389 in Mechanics | Relativity for Bossman

Question

The body of mass 5kg is suspended by an inextensible string from a nail 0 and is pulled by a horizontal force F until the angle of inclination of the string to the vertical is 60 degrees. Calculate the value of F and the tension in the string( g as 10m/s

Answer 1

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

$\begin{cases} -F+Tsin\alpha=0 \\ -mg+Tcos\alpha=0 \end{cases}$

$\begin{cases} T=\frac{mg}{cos\alpha} \\ F=Tsin\alpha \end{cases}$

$\begin{cases} T=\frac{mg}{cos\alpha} \\ F=mgtan\alpha \end{cases}$

$T=\frac{5 \cdot 10}{0.5}=100$ (N)

$F=5\cdot10\cdot1.73=86.5$ (N).

Answer: 86.5 N, 100 N.