[mrose17]

2 men are carrying a 200 kg box; one at each end. They are on stairs which are at a 45 degree angle with the ground. The box they are carrying is also at a 45 degree angle, so the stairs and ground are parallel. The lengths of the box are 1.23 m and .5 m, and both men exert a vertical force on the box. They aren't moving/there's no velocity or acceleration involved. How do you calculate the force each man applies to the box?

[ebaines]

The only forces acting on the box are the support by the two men, which must counteract the weight of the box. You will need to know where the center of mass is for the box, then you can sum the torques around either end of the box and set that to zero - this will give you the force applied by the man at the other end of the box. If you assume that each man's supporting force is purely vertical, and that neither applies any torque of their own, then for sum of torques about point 1 you get:



If you do the math you should find that the angle of the box to the ground doesn't matter. Post back with your answer and we'll check it for you.

[mrose17]

Thank you! Your answer was a huge help! For F1 I got 1372 N and for F2 I got 588 N.
In the original question I had one of the values wrong; one side of the box is actually 1.25 m.

[ebaines]

Hmm... where is the center of mass for the box? If it's in the middle then I disagree with your answer. Suppose the box was held parallel to the ground instead of at 45 degrees - what would you get then?

[mrose17]

I calculated it with the center of mass at the center of the box, and the men are on stairs at a forty five degree angle so the box and stairs are parallel. Do I need to get rid of the sin45 and tan45 I used in my equation?

[ebaines]

Not sure why you are using sines and tangents. From the figure I posted earlier:

Sum of torques about point 1 = W*L/2*cos(45) - F2*L*cos(45) = 0
so F2 = W/2.

[mrose17]

I used sine and tangent in calculating the lever arm for the weight of the box. I thought since the weight and the forces aren't all in a line, the lever arm isn't Lcg*cos45 anymore. So I ended up getting sin45*(.625-.25/tan45) for the lever arm for W

[ebaines]

I see you point - I left out the fact that the center of mass is not in line with the points where the men are supporting the box. So the lever arm from point 1 to the center of mass is actually:

(L/2)cos45 - (H/2) sin45

I think you have the sine and tan terms reversed, but it doesn't matter because sin45 = cos45.

One last point - the lever arm from point 1 to where the 2nd man is holding the box is L*cos(45). I think you may have used something else? I get F1 = 1169N and F2 = 791N.

[mrose17]

Thank you!