# Torque and forces (Beginner level)

#### ZimmerJ

Joined Dec 9, 2020
58
Hello, i have recently started learning about torque (Nm) and i have encountered my first problem.

As described in the included picture, i want to figure out how one would calculate the forces that point H (hinge) and the string in the other end are subjected to in case 3.

I did some simple calculations as examples, just to show how i understand things so far:

- When a one-dimensional line is horizontal, like in case 1, calculating the forces at those points that "carries" the line seems straight forward.

- On to case 2, suppose the line is balanced vertically, point H will be subjected to the whole weight of 50 N and therefore i excluded the string.

- Now, having a tilted line that is held up by point H and the string in the other end, as in case 3, i just can't seem to figure out how to calculate the forces that these ends are subjected to?

I thought of using the same principle for calculating the force exerted on the string (Fb), as when calculating F1 and F2 using the torque-value, but it will always result in half the total weight → 25 N, no matter the tilt (Intuitively i figured this has to do with using the point mass in the torque calculation but i'm not sure why, and this is probably the deeper and perhaps mathematical knowledge that i am lacking).

So how do you go about it? To me it is obviously more force exerted on point H, until the line is tilted all the way down to a horizontal position as in case 1, where forces are equally distributed between the two ends.

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#### hrs

Joined Jun 13, 2014
414
In case 1, how do you get from 50 N - Fb = Fb * L2/L1 to Fb = Fa?
If a CW moment is positive then a CCW moment is negative. Be explicit about which point you resolve the sum of the moments.

In case 3 F2 = 0. Just sum the forces in y (vertical) and the moments around H.

#### ZimmerJ

Joined Dec 9, 2020
58
In case 1, how do you get from 50 N - Fb = Fb * L2/L1 to Fb = Fa?
If a CW moment is positive then a CCW moment is negative. Be explicit about which point you resolve the sum of the moments.

In case 3 F2 = 0. Just sum the forces in y (vertical) and the moments around H.
I skipped a little bit there. For that equation, it only leads to Fb = 25 N. Since the distance from the center to Fa is equal to the distance between the center and Fb, Fa also equals 25 N. Hence, Fb = Fa = 25 N. My bad.

If Fb should be considered negative, sure. But what i am interested in is the net supporting force for point H and String respectfully. In case 1, they are of the same magnitude.

Force F1 and F2 in case 3 i understand are misleading. It was only to display the supporting force in case of a diagonal string at different angles. What i want to know in case 3, is the magnitude of the supporting force in the vertical string.

Thanks.