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Below is the homework problem I am working on:

A 18.3-kg object hangs in equilibrium at the end of a rope (taken as massless) while a wind pushes the object with a 63.9-N horizontal force. Find the tension in the rope and the rope's angle from the vertical. The acceleration due to gravity is $9.81 \space m/s^2$.

I am having a hard time correctly defining the system of two equations to solve for tension and angle. I came up with $$T\cos{(\theta)} -F_H=0 \\T\sin{(\theta)}-(-F_g)=0$$ where $F_g$ is the force due to gravity and $F_H$ is the horizontal force. I use a free body diagram where $F_H$ is on the x-axis and $F_g$ is on the y-axis. Apparently, my setup is incorrect. Could you put me on the right track. Let me know if more clarification is needed.

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  • $\begingroup$ what makes you say your setup is incorrect? $\endgroup$
    – Brian Moths
    Commented Apr 29, 2014 at 3:00
  • $\begingroup$ Does the wind decrease tension on the rope? $\endgroup$
    – LDC3
    Commented Apr 29, 2014 at 3:08
  • $\begingroup$ @NowIGetToLearnWhatAHeadIs I enter my answers into the check box and it says they are incorrect. $\endgroup$
    – Koba
    Commented Apr 29, 2014 at 3:08
  • $\begingroup$ @LDC3 Thats the whole problem description. I assume it does not because this a problem from a 101 course we are only in the beginning of it. $\endgroup$
    – Koba
    Commented Apr 29, 2014 at 3:09
  • $\begingroup$ No, I'm saying it looks like your equation is decreasing the tension on the rope instead of increasing it. If there is no wind, the tension is due to gravity alone. When there is a wind, the tension increases. So, is your equation correct, or do you have the wrong sign. $\endgroup$
    – LDC3
    Commented Apr 29, 2014 at 3:23

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Assuming $\theta$ is the angle measured from the vertical, then $Tcos(\theta)$ will give you the vertical component of the tension, whereas $Tsin(\theta)$ will give you the horizontal component. Therefore, the expression $Tsin(\theta)$ should be equal and opposite to $F_H$, and $Tcos(\theta)$ should be equal and opposite to $F_g$. I believe your system has this backwards.

Also depending on exactly what you are plugging in for $F_g$ (meaning, whether you are plugging the absolute value of the gravitational force or not), this could be causing you problems with the signs you are using. Personally, I would write the second equation like this: $$ T_y-F_g=0$$ Then plug in the magnitude of the gravitational force for $F_g$.

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  • $\begingroup$ Oh yeah. My system has it backwards. Thanks for pointing out. $\endgroup$
    – Koba
    Commented Apr 29, 2014 at 16:57

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