Now, using the instantaneous axis, I get that net torque = mg * r - (t * 2r)
= 980 * 0.4 - 0.8(30)(9.8 + a)
= 392 - 235.2 - 24a
= 156.8 - 24a
Which is again equal to I *(a/r) which is 24 * a/0.4 = 60a
so 156.8 = 84a
a = 156.8/84 = 1.86666667 again (is this the answer?)
Also, on a quick side...
Just realized I forgot that for the barrel, Fnet = 100 * a (and a is the same for both masses)
So my new revised process is fs = Fg - Fnet - T
= (100)*(9.8) - 100a - 30(9.8 + a)
=980 - 294 - 100a - 30a
= 686-130a
so net torque = (fs * r) - (T * r)
= 0.4(686-130a) - 0.4* 30(9.8 + a)
=...
So basically for the FBD of the big barrel, I've got Fnet = Fg - (T+fs) where fs is friction. From this I get:
fs = Fg - T = (100)(9.8) - 30(9.8 + a)
Now I know torque = fs * radius, and that tension and friction work in opposite directions.
I know it sipns counter clockwise, so:
net...
Homework Statement
A barrel is lowered into a boat using the illustrated apparatus. The barrel can be considered to be a uniform cylinder with M=100kg and r=0.40m. The weight on the other end of the rope has m=30kg. Assume that the barrel does not slip against the wall, that the other 2...