Find the forces applied to a rotating level

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melissafern
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Homework Statement
Solve parametrically the system with g, load cell measurements and know accelerations.
Relevant Equations
to be found
Homework Statement: Solve parametrically the system with g, load cell measurements and know accelerations.
Homework Equations: to be found

Consider that we know angles tetha(t) of a mass-less link rotating about its centerr O. - so we also have velocities and accelerations - and values of bidirectional forces on a load cell under the two masses. So we have the tengential and normal forces applied to the 2 segment extremities. I need to find the two forces vector (4 unknown) which are the 2 forces applied over the masses.
 

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Welcome to the PF. :smile:

Please go ahead and start the problem so we can offer tutorial help. Until you show us your work, we are not allowed to help you.

What equations do you think you should start with? And what is the significance of the object being a "level"? I don't get that part of the question yet...
 
I have a massless and non-deformable bar with a central point which is fixed O (geometrical midpoint) and that allows rotation of the bar and separates completely tensions along the bar (so that the tangential forces on the right side are discharged in the center as well as the left side ones). On each side there is a mass ad a force acting on it. The two mass are resting on a load-cell (as a scale all around it = the red rectangular shape in the drawing) that measure the resultant forces along and perpendicular to the bar.

Input values are M1, M2, (W1)tang, (W1)norm, (W2)tang, (W2)norm, the length of the bar, angle (inclination of the bar) as a function of t and its derivatives angular velocity and angular acceleration alpha. The unknown are the 2 force vectors F1 and F2.
I would also like to ask if the problem is solvable with alpha among the unknowns.
Gravity is in the vertical direction of the drawing.

1. (W1)n= M1ωr+F1sinβ1+ M1gsinθ

2. (W2)n= M2ωr+F2sinβ2+ M2gsinθ

3. (W1)t= M1gcosθ+F1cosβ1-M1αr

4. (W2)t= M2gcosθ+F2cosβ2-M2αr

5. F1cosβ1r+M1gcosθ-M1r2α= F2cosβ2r+M2gcosθ-M2r2α

dθ/dt=ω

dω/dt=α
 

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