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smayya337 wrote: ↑April 25th, 2020, 7:49 pm

Umaroth wrote: ↑April 25th, 2020, 12:47 pm 0.0407, less than 1 because it's the axle doing to driving

A door is 1.00 m wide and the handle is 0.75 m from the hinge. What is its IMA?

1.33, since it should just be 1.00/0.75

I'll post a new question tomorrow.

Actually, that's incorrect, it should be 0.75/0.50= 1.5 because the door's center of mass is in the middle, 0.50 m from the hinge.

Umaroth wrote: ↑April 25th, 2020, 8:04 pm

smayya337 wrote: ↑April 25th, 2020, 7:49 pm

Umaroth wrote: ↑April 25th, 2020, 12:47 pm 0.0407, less than 1 because it's the axle doing to driving

A door is 1.00 m wide and the handle is 0.75 m from the hinge. What is its IMA?

1.33, since it should just be 1.00/0.75

I'll post a new question tomorrow.

Actually, that's incorrect, it should be 0.75/0.50= 1.5 because the door's center of mass is in the middle, 0.50 m from the hinge.

Oops, you're right! I should probably get back into practicing for Machines...

On a certain Atwood machine, blocks A and B are on opposite ends of a rope that is connected to a fixed pulley. Assume that the masses of the rope and pulley are negligible, and that there is no friction. If block A has a mass of 6.0 kg and block B has a mass of 2.0 kg, what is the absolute value of the acceleration of block A? (I don't really care about the sign, hence the absolute value.)

Alright, here's my attempt:
The net force on block A is 39.2N (weight of A - weight of B), and divided by the mass, the acceleration is 6.5 m/s^2

A 40.0cm gear drives a 20.0cm gear.
a. If the input torque is 10.0N*m, what is the output torque?
b. If the input rotational velocity is 30.0rpm, what is the output rotational velocity?
c. A 30.0cm gear is added in between these 2 gears. Now what is the output velocity if the input velocity is 30.0rpm?

Kinda simple, but I'm not super creative at the moment.

a) still 10.0 N*m, torque stays the same, I think you meant output force
b) -60.0 rpm, negative implying opposite direction
c) +60.0 rpm, same direction

A bird is perched on the middle of a long string. Is there any way to apply tension to the ends of the string such that the string remains completely taut? Explain.

Umaroth wrote: ↑April 27th, 2020, 9:46 am a) still 10.0 N*m, torque stays the same, I think you meant output force
b) -60.0 rpm, negative implying opposite direction
c) +60.0 rpm, same direction

A bird is perched on the middle of a long string. Is there any way to apply tension to the ends of the string such that the string remains completely taut? Explain.

Yeah, thanks for catching that torque error.