Solving problem concerning simple machines

Question

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!

score 0 · Accepted Answer · answered May 20 '18 at 17:48

0

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

$$ f_0 =49.9433\\ h = 39.8865 $$

answered May 20 '18 at 17:48

Cesareo

I really don't know what you did there. – Busi May 20 '18 at 18:42
$h$ and $f_0$ are the module forces involved. No more explanations. – Cesareo May 20 '18 at 18:48
They're wrong, you know? – Busi May 20 '18 at 18:55
If you say that, show me the solution then. – Cesareo May 20 '18 at 19:00
My teachers found this answer. However, should I have mentioned them $\sin 37 = 0.6$, $\cos = 0.8$? – Busi May 20 '18 at 20:09
$\sin(37^{\circ}) = 0.601815$ and $\cos(37^{\circ}) = 0.798636$ – Cesareo May 20 '18 at 20:15
Can we say that $\dfrac{80}{\sin 37 + \cos 37}$? – Busi May 21 '18 at 10:26

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