enter image description here What would be the force F to keep the rod AB in equilibrium? I got confused while using Lami's theorem. $$\frac{F}{\sin160^\circ}=\frac{F_1}{\sin60^\circ}=\frac{W}{\sin140^\circ}$$ If F is keeping it in equilibrium, I can also write, $F^2=W^2+F_1^2+2WF_1\cos160^\circ$

Please give me the hint where am I doing wrong? Have I done any mistake in angle calculation? Thanks in advance.

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    $\begingroup$ Shouldn't this be in PhysicsSE? $\endgroup$ Mar 14, 2019 at 2:29
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    $\begingroup$ Does $F$ act perpendicular to the rod? If so, your angles are wrong.$$\frac{F}{\sin160^\circ}=\frac{F_1}{\sin40^\circ}=\frac{W}{\sin120^\circ}$$ $\endgroup$ Mar 14, 2019 at 2:49
  • $\begingroup$ @ShubhamJohri I have confusion with the direction of F too. It's not specified in the description of the problem. How did you make 40 and 120? they don't even sum up 360. $\endgroup$
    – raf
    Mar 14, 2019 at 3:24
  • $\begingroup$ Ignore my earlier comment. The figure misled me into thinking that $F$ acts perpendicular to the rod. In fact, for the rod to be in equilibrium, you don't need to be given the direction of $F$ since $F$ must act opposite (and have magintude equal) to the resultant of $W$ and $F_1$. You can calculate the magnitude of $F$ using the triangle law and, assuming the angle between $F$ and the vertical is $\beta$, use the sine rule equality$$\frac F{\sin160^\circ}=\frac{F_1}{\sin\beta}$$to find $\beta$. I got the answer $F=45.51N$ and $\beta=48.72^\circ$. $\endgroup$ Mar 14, 2019 at 12:46
  • $\begingroup$ Depending on what angle you took $\beta$ as, you might get the supplement of what I got. $\endgroup$ Mar 14, 2019 at 12:54


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