Why Is My Calculated Torque Different from the Textbook's Answer?

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The discussion revolves around calculating the gravitational torque of a uniform rod, with a specific focus on a pivot point located 25 cm from one end. The user initially calculated the torque as 14.7 N·m, while the textbook states it should be -4.9 N·m. Participants suggest dividing the rod into separate masses to compute the sum of their torques and emphasize using the center of mass for a more straightforward calculation. Drawing a diagram to visualize the rod's center of mass and the pivot point is recommended to clarify the distance for torque calculation. Understanding the distribution of mass and the correct application of torque formulas is essential for resolving the discrepancy.
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Hello all, I've been wrestling with this question for awhile, please help!

There is a 2.0kg, uniform, horizontal rod measuring a total of 100cm. Toward the left end of the rod there is a point 25cm from the left end and 75cm from the right end. The question is to find the gravitation torque about the point.

I used the formula T = rF and got 14.7 N . m

the book says the answer is -4.9 N . m

im so confused, please help!
 
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You need to divide the rod into separate masses and calculate the sum of their torques.
 


hello

i considered separating the masses but i didnt know of any feasible means of doing so
 


bigsaucy said:
hello

i considered separating the masses but i didnt know of any feasible means of doing so

You mean you can't determine 1/4 and 3/4 of 2.0kg?

A more expedient route to the torque when the rod is of uniform density is to simply use the distance of the center of mass of the rod (as a whole) from the point about which the torque is to be found.
 


i'm so terribly confused =[
 


Draw a diagram of the rod. Pencil in the rod's center of mass. Pencil in the location of the pivot point (point about which you are trying to find the torque due to gravity). What's the distance between the two points?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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