Torque TBR problem (easy?)

[ilovemedi]

---

Members don't see this ad.

Question: A long plank of mass m and Length L lies on the ground. You wish to pick it up at one end, leaving the other at the ground. What minimum force must you apply?

A) mg/2
B) mg.
Answer: A
Explanation: END SERVES AS PIVOT POINT. TO PICK UP, THE TORQUES ABOUT END MUST BE EQUAL (sorry for caps). Two torques act on the plank: one form the weight, of plank, and one from you picking up the plank. The gravitational force on the plank is half the plank length (L) away from the stationary end, and the force (F) you apply is L away from the stationary end. Setting these two torques equal is L/2mg=LF. Solve for F.

I get it intuitively. if you only lift half of an object, less force required. But there reasoning confuses me so much. I thought we IGNORE the plank/rod/whatever the weights are balanced on unless it tells us to. Second, since when is gravitation force on the plank half the plank length?! WTH does that even mean?

 

[brood910]

You ignore the mass of rod/plank when the pivot point is at the center.
Since we set the pivot point at the end in this case, you have to consider its mass.

Also, their weight is at the center. ALWAYS. So, from the center to the end = L/2.

 

[Blueprint MCAT Tutor]

Brood is right, and we may even say more generally:

In rotational equilibrium problems where gravity is providing the force, treat all objects as if they only existed at a single point - their center of gravity.