Question

If these components have weights WA = 50000 lb , WB=8000lb, and WC=6000lb, determine the normal reactions of the wheels D, E, and F on the ground.

Answer 1

Answer:

F(F) = 15037 lb

F(E) = 24481.5 lb

F(D) =  24481.5 lb

Explanation:

(The diagram of the figure and Free Body Diagram is attached)

Data given:

W(A) = 50,000 lb

W(B) = 8000 lb

W(C) = 6000 lb

∑F = 0

F(F) + F(E) + F(D) - W(A) - W(B) - W(C) = 0

F(F) + F(E) + F(D) = W(A) + W(B) + W(C)

F(F) + F(E) + F(D) = 50000 + 8000 + 6000

F(F) + F(E) + F(D) = 64000 lb

∑M(o)

∑M(o) = M(F) + M(E) + M(D) + M(A) + M(B) + M(C)

Where

M(F) = 27i × F(F)k = -27F(F)j

M(E) = 14j × F(E)k = 14F(E)i

M(D) = -14j × F(D)k = -14F(D)i

M(A) = 7i × -50000k = 350,000j

M(B) = (4i - 6j) × -8000k = 48000i + 32000j

M(C) = (4i + 8j) × -6000k = -48000i + 24000j

∑M(x) = ∑M(i) = 0

∑M(i) = 14F(E) - 14F(D) = 0

F(E) = F(D)

∑M(y) = ∑M(j) = 0

∑M(j) = -27F(F) + 350,000 + 32,000 + 24,000 = 0

27F(F) = 406,000

F(F) = 15037 lb

F(F) + F(E) + F(D) = 64000 lb

F(E) = F(D)

F(F) + 2F(E) = 64000

2F(E) = 64000 - 15037

2F(E) = 48963

F(E) = 24481.5 lb

F(D) =  24481.5 lb