# Thread: moments-with forces applied at an angle.

1. ## moments-with forces applied at an angle.

The bar shown is hinged at H and it is in equilibrium. Find the angle $\displaystyle \theta$

I really don't know were to start, how do I find the angle when i don't know the value of any other sides except for the hypotenuse?

2. Hi

The 20N force can be separated into 2 parts : 20 cos(theta) i + 20 sin(theta)j
The projection on x axis (20 cos(theta)) gives no momentum with respect to H
The projection on y axis (20 sin(theta)) gives a momentum with respect to H equal to 20 sin(theta) x 5
The bar being at equilibrium, this momentum must be equal to the one generated by the 15N force, which is 15 x 1
20 sin(theta) x 5 = 15 x 1
sin(theta) = 3/20 = 0.15
theta = 8.6 deg

3. Let pivot be at H. Therefore:
$\displaystyle \tau_{\text{clockwise}}=\tau_{\text{anticlockwise} }$

You may have forgotten this if you dealt with perpendicular forces mostly, so recall that $\displaystyle \tau = F\cdot d\sin{\theta}$. Therefore:
$\displaystyle 15 = (20)(5)\sin{\theta}$

Edit: running beat me to it. Although with simple problems like these, I think it's better to deal with torques as scalars.