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
$F_{T}$	$F_{T}\cos(60^{\circ })$	$F_{T}\sin(60^{\circ })$
$\Sigma F$	0	0

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

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

$F_{T}={78.4 \over \cos(60^{\circ })}={78.4 \over 0.5}=157N$

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

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

$m={F_{T}\sin(60^{\circ }) \over g}={0.87F_{T} \over g}={0.87(157) \over (10)}=13.7kg$