- An object with the weight of $80N$ is balanced with the force of $\vec {F}$ as seen on the diagram. What is the value of $\vec {F}$? (Ignore the weight of pulley and friction)
Resolving the components in $y$-axis.
The $y$-axis component of $F$ is $F\sin (37)$. It is balanced by the weight $80 N$.
$$F\sin(37) = 80$$
$$F = \dfrac{80}{\sin(37)} = \dfrac{80}{0.6} = \boxed {133.33 N}$$
However, there's no upward option of $50N$. Am I wrong?
Regards!
The intervening forces are
$$ \vec f_1 = f_0(0,1)\\ \vec f_2 = f_0(\cos(37\pi/180),\sin(37\pi/180))\\ \vec f_3 = h(-1,0)\\ \vec f_4 = 80(0,-1) $$
and for equilibrium
$$ \vec f_1+\vec f_2+\vec f_3+\vec f_4 = 0 $$
giving
$$ f_0 =49.9433\\ h = 39.8865 $$