A 20.0-m-long uniform beam weighing 690 N rests on walls A and B, as shown in th

Question

A 20.0-m-long uniform beam weighing 690 N rests on walls A and B, as shown in the figure (Figure 1)

AC=3m AB=12m BD=5m

I know that the max weight of someone to walk to the end of D without tipping is 690 N. Also that wall A's force is 0 N when standing on D while B's force is 1400 N when standing on D.

Find the forces that the walls A and B exert on the beam when the person is standing at a point 1.7 m to the right of B?

I keep getting 230>x>220 for A and 1170>x>1200 for B which is incorrect. If you could solve and express how you got the answer that would be excellent. Also,

Find the forces that the walls A and B exert on the beam when the person is standing 1.6 m to the right of A?

Thank you!

Explanation / Answer

this is little bit confusing

When 690 N person stands 1.7 m to the right of B then again taking point B as the pivot point:

Torques = 1.7(690) + 12(FA) = 1580 + 12(FA)

Torques = 5(690) = 3450N

1173 + 12(FA) = 3450

12(FA) = 3450 - 1173 = 2277N

FA = 2277/12 = 189 N

find FB use STATIC equilibrium of forces:

FA + FB = 690+690 = 1380

FB = 1380 - FA = 1380 - 189 = 1191 N

for part (d) repeat same process as in (c)

with a new sketch with W, 1.6 m to the right of A and using A as the pivot point

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