# Finding the magnitude of F An object with the weight of $80N$ is balanced with the force of $F$ as seen on the diagram. What is the magnitude of $F$? (Ignore the weight of pulley and friction)

$\sum F_y = 0$

$F_y \sin 37 - 80 = 0 \implies F_y \sin 37 = 80 \implies F_y = 133.33N$

$\sum F_x = 0$

$F_x \cos 37 = 0 \implies F_x = 0$

Finally, we have

$F = 133.33N$

Am I right?

There are two sides of the cable around the pulley that helps to support the 80 N load. See the diagrams below. So the vertical force from the cable around the pulley is $F+F\sin(37)$, not $F_y\sin(37)$. This gives the sum of vertical forces as $F+F\sin(37)-80=0$ from which F=49.9 N. (Note that $F_y\sin(37)$ does not make sense.)

The horizontal force from the cable around the pulley is $F\cos(37)$, not $F_x\cos(37)$. Furthermore, there is a force in the horizontal "rope" of magnitude T, so the sum of forces in the horizontal forces is $F\cos(37)-T=0$ from which T=39.9N.

Here is a free body diagram showing all of the forces acting on the pulley. The statement to neglect the friction in the pulley indicates that the tension in the cable is the same value, F, on each side of the pulley.

• Does the question seem incorrect? because everyone finds anohter answer.
– Busi
May 22 '18 at 17:35
• If two people get a different answer to the same question, I think the problem is not with the question. ;-) If my free body diagram is reasonable, then your answer of F=133 N indicates that it takes somewhere between 133 N (if just one F holding the pulley) to maybe as much as 266 N (two F's because you are saying the angle is really 90 degrees since Fx=0) to hold up an 80 N weight. Does that seem correct? May 22 '18 at 21:52
• Yes, However. 19 people found the answer 57.12N or 133.33N. So, am so scared.
– Busi
May 23 '18 at 5:33
• I agree with this answer. If Fx=0 angle woold be 90 degree, not 37 degree. May 23 '18 at 18:46
• @Katarina Which answer seems right? or is there anything going wrong with the question?
– Busi
May 24 '18 at 18:07