Newtons Laws EX 26

Since the bob is held in place, its acceleration is equal to zero. There are three forces acting on the bob: the force exerted by the spring $F_S$, the force exerted by the string $F_T$ and the force of gravity $F_G$. Assume that the mass of the bob is $m$ and then $F_G=mg$. We choose the coordinate system with the horizontal x-axis and draw the free body diagram. Given that the string makes a 30° angle with the vertical, the vector $F_T$ makes a 60° angle with the x-axis.

Next we compute the components of all forces. We begin by computing the magnitude of $\overrightarrow{F_S}$:

$F_S=kx=(3.92\times 10^3)(2\times 10^{-2})=78.4N$

Forces	x-component	y-component
$F_S$	-78.4 N	0
$F_G$	0	-mg
$F_T$	$F_T\cos (60^{\circ })$	$F_T\sin (60^{\circ })$
$\mathrm{\Sigma }F$	0	0

The sum of the forces in the x-direction is zero:

$-78.4+F_T\cos (60^{\circ })=0$

$F_T=\frac{78.4}{\cos (60^{\circ })}=\frac{78.4}{0.5}=157N$

The sum of the forces in the y-direction is zero:

$F_T\sin (60^{\circ })-mg=0$

$m=\frac{F_T\sin (60^{\circ })}{g}=\frac{0.87F_T}{g}=\frac{0.87(157)}{(10)}=13.7kg$