Part A Find the maximum weight of a person who can walk to the extreme end D wit

Question

Part A

Find the maximum weight of a person who can walk to the extreme end D without tipping the beam. (Ans in N)

Part B

Find the force that the wall A exert on the beam when the person is standing at D.

Express your answer using two significant figures. (Ans in N)

Part C

Find the force that the wall B exert on the beam when the person is standing at D.

Express your answer using two significant figures.(Ans in N)

Part D

Find the force that the wall A exert on the beam when the person is standing at a point 2.0 m to the right of B.

Express your answer using two significant figures.(Ans in N)

Part E

Find the force that the wall B exert on the beam when the person is standing at a point 2.0 m to the right of B.

Express your answer using two significant figures.(Ans in N)

Part F

Find the force that the wall A exert on the beam when the person is standing 2.0 m to the right of A.

Express your answer using two significant figures.(Ans in N)

Part G

Find the force that the wall B exert on the beam when the person is standing 2.0 m to the right of A.

Express your answer using two significant figures.(Ans in N)

If you decide to write out and post a pic of work, please make it legible so that I can follow it, Thanks!

Explanation / Answer

(1) M = 0 = W * 5m - 820N * 5m
(since at the tipping point, the force on the left wall is zero).
Then max weight W = 820 N

(2) As mentioned, Fa = 0 N. Fb = 820N + 820N = 1640 N
(vertical forces must sum to zero)

(3) Let's sum the moments about A:
M = 0 = Fb * 12m - 820N * 14m - 820N * 7m Fb = 1435 N
Therefore Fa = 820N + 820N - 1435N = 205 N

(4) Again, let's sum the moments about A:
M = 0 = Fb * 12m - 820N * 2m - 820N * 7m Fb = 615 N
Therefore Fa = 820N + 820N - 615N = 1025 N

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