Beam and cable problem: tension in the cable using torque and hinge force at the end of the beam.

[@tlotlisogladwinmore3603]

You just made them look easy. Thank you.

[@ZaksLab]

Thanks! -- Zak

[@gad3386]

Thank you!

[@ZaksLab]

You're welcome! -- Zak

[@dabadababa]

"Hello, I have a question regarding the beam and cable problem in this video. At around 1:12, you mentioned that the torque at the connection point of the cable and the beam is zero when considering that point as the rotation axis. However, since there is an angle between the tension cable and the beam, shouldn't there be a torque component due to the tension force? I would appreciate if you could clarify this point. Thank you!"

[@ZaksLab]

If you take the connection point of the cable as the rotation axis, then the tension exerts no torque around that rotation axis, because the lever arm is zero! There are torques about that point exerted by the force of gravity at the center of mass and the hinge force because those forces both act at some distance from the chosen rotation axis. When we choose a different rotation axis, say the hinge, then the tension will exert a CCW torque with respect to that axis because it now has a non-zero lever arm. And the main idea with torque analysis in static equilibrium is that you can choose rotation axes anywhere you want for the analysis: the net torque about any chosen axis is zero because the system is not moving at all! Hope this helps. z

[@robertsakatay8579]

Hello, why do we neglect the tension exerted by the cable at the end of the cable in which the cable is being mounted onto the wall?

[@ZaksLab]

The key here is that we're analyzing all forces and torques *acting on the beam*. We have included the force/torque exerted on the beam by the cable, and the force exerted on the wall by the cable is not part of the analysis of forces/torques on the beam. z

[@jun-hsuangjen7385]

Nice video! I am wondering whether the Fx of hinge is the normal force and the Fy is the frictional force ?

[@ZaksLab]

Not if it's supported by a hinge, but if it was held in place by friction I suppose that's what the two components would be. -- Zak

[@jun-hsuangjen7385]

@@ZaksLab so if it’s supported by the hinge, then what are those two components called? Thank you so much for your reply!

[@ZaksLab]

@@jun-hsuangjen7385 I'm not aware of a special name for them --z

[@blueming1246]

I don't get it, how did you get the 1670N

[@ZaksLab]

solving for T = 120*9.8*1.5/(sin(25)*2.5)

[@blueming1246]

@@ZaksLab thank you

[@joelsenpai5141]

Why is there a 1.5 meter mark

[@ZaksLab]

To compute the torque exerted by gravity, we look at the force of gravity acting at the center of mass of the rod. Since the rod is uniform and 3m in length, the center of mass is the 1.5m mark. -- Zak

[@KrisdaCannotbrain]

How come torque is equal to tension? May u rationalize plz

[@ZaksLab]

Careful - you can't say torque is equal to tension (one is a torque and one is a force!) There are two torques relative to the hinge. Mg multiplied by lever arm going clockwise, then the perpendicular component of tension Tsin(theta) multiplied by its lever arm going counterclockwise. In each case the torque is given by the perpendicular component of the force multiplied by the lever arm (distance to the rotation axis). z

[@hazelnut3443]

Hye , for the calculation check , i got slightly different which the tension i a bit more than the Force at hinge ? Am i correct ??? Cause ive done many calculation but still dont get the exact same amount

[@ZaksLab]

If you're just off in the last digit, that's probably me doing too much rounding error. Give me your numbers, and I can check against them. -- Zak

[@hazelnut3443]

@@ZaksLab for the counterclockwise, i got 705.77N while for the clockwise is 704.78N

[@ZaksLab]

@@hazelnut3443 It has to just be the rounding. I kept a bunch of extra digits and re-did all the calculations, and the two torques for the double check both come out to 705.6N. You're close enough where I don't think you made any concept errors, it's just an artifact of the crude rounding early in the problem. z

[@hazelnut3443]

@@ZaksLab i see , thank you for your reply ! Its my assignment and I’m kind of confused with this chapter , your explanation help me so much !

[@firdausbasri7237]

What if the slings is more than 3

[@ZaksLab]

If you need more equations to evaluate more unknowns, you can just generate more equations by evaluating the net torque about different axes. You can go to my 'about' page and contact me by e-mail if you have a specific question in mind, and I'll check it out. -- Zak

[@SI_Sourav_Mondal]

The distance of line of action of mg from the axis of rotation should be 1m not 1.5m.... Sir pls check it...

[@ZaksLab]

the beam is 3m long, making the center of mass 1.5m from the left end which was acting as the rotation axis for the analysis -- z

[@SI_Sourav_Mondal]

@@ZaksLab ok sir... I have got it... I thought the AOR is at right end where u have marked....

[@ZaksLab]

@@SI_Sourav_Mondal in the first part of the video I did a qualitative analysis about that axis to establish the correct direction for the y component of the hinge force, then I switched to using the hinge as the rotation axis for the quantitative torque analysis-z