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.
Best Answer
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 $F_g$.