Torque and forces (Beginner level)

Thread Starter

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.

Any answers are appreciated. Thanks
 

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hrs

Joined Jun 13, 2014
359
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.
 

Thread Starter

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.
 

Thread Starter

ZimmerJ

Joined Dec 9, 2020
58
I've have come to the conclusion that i am mixing rotational movement with the force of gravity. It now makes sense.
 
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