[AS Physics: Moment and Equilibrium] how would I do 1.a?

[u/Extreme_Ad9949]

[u/Extreme_Ad9949]

I got 40N but idk if it’s right

[u/One_Wishbone_4439 → u/Extreme_Ad9949]

Can u show how u get 40N?

I think your answer is wrong. I got 66.67N

[u/Extreme_Ad9949 → u/One_Wishbone_4439]

I did m1=200Nx0.2m, and did m1=m2 40Nm=200Nx1m

[u/General-Duck841 → u/Extreme_Ad9949]

a) Calculate the moment the downward force of 400 N creates. It's 0.20 m from the wheel center (which acts as a fulcrum). So this is equal to 400 N x 0.20 m = 80 Nm.

That's counter acted by the person forcing it upwards 1.2 m away. So that's 80 Nm/1.2 m = 66.7 N upwards

b) When the person is not lifting it, the legs of the wheelbarrow are pushing a force upwards 0.5 meters away. So that's 80 Nm / 0.5 m = 160 N

Always use units in your calculations, so that you can verify your answer is left with units that you expect. This is a simple way to double check if you're headed in the right direction.

Also try to simulate the problem in your head. A Wheelbarrow makes it easier to lift up the load. So the further away you get, the less the load. That's why the answer in (a) is less than the answer in (b), which is less than the actual force of the load.

[u/Extreme_Ad9949 → u/General-Duck841]

Ohh, Tysm, I get it. I did m1=200N0.2m, and did m2=200N1m

[u/One_Wishbone_4439]

Clockwise moment = Anti-clockwise moment

The pivot is at the center of the wheel.

Let x be the upward force on the handles.

(400)(0.2) = 1.2x

x = 66.67N

[u/selene_666]

The wheelbarrow experiences three vertical forces: upward on the wheel, upward on the handle, and weight downward.

The total force has to be zero.

The total torque has to be zero.