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A uniform steel beam has a mass of 1800 kg. on it is resting half of an identical beam, as shown in fig. 9-44. what is the vertical support force at each end? left end n (upward) right end n (upward)

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shinmin
Let the left end A and the right end B. 
Let w = the weight of the full beam. 
0 = -(w/2)*(L/4) - (w)*(L/2) + Fa*L 
Fa = [(w/2)*(L/4) + (w)*(L/2)]/L = w/8 + w/2 = 5/8*w = 5/8*m*g = 5/8*1800*9.81 Fa = 11036.25 N 
Fa + Fb = w Fb = w - Fa = 1.8*(1800*9.81) - 11036.25 Fb = 20748.15 N

Answer:

Support force at side A: 15,435 N

Support force at side B: 11,025 N

Explanation:

Part A:

Because we are looking for the force at side A, we set [tex]F_B=0[/tex] for easier calculations.

Recall that:

[tex]\sum\tau=0; \tau=F_\perp*r; W=m*g[/tex]

And given that:

[tex]W_B=\frac{1}{2}W_A[/tex]

Where [tex]W_B[/tex] is the weight of the smaller mass.

Hence:

[tex](-W_B*\frac{3L}{4})-(W_A*\frac{L}{2})+(F_A*L)=0\\\\\frac{(W_B*\frac{3L}{4}) + (W_A*\frac{L}{2})}{L}=F_A[/tex]

Substitute [tex]\frac{1}{2}W_A[/tex] for [tex]W_B[/tex]

[tex]\frac{3W_B}{4}+\frac{W_A}{2}=F_A\\\\\frac{3W_A}{8}+\frac{4W_A}{8}=F_A\\\\\frac{7}{8}W_A=F_A\\\\F_A=\frac{7}{8}*1800*9.8=15,435 N[/tex]

Part B:

Because we are looking for the force at side B, we set [tex]F_A=0[/tex] for easier calculations.

Recall that:

[tex]\sum\tau=0; \tau=F_\perp*r; W=m*g[/tex]

And given that:

[tex]W_B=\frac{1}{2}W_A[/tex]

Where [tex]W_B[/tex] is the weight of the smaller mass.

Hence:

[tex](-W_B*\frac{L}{4})-(W_A*\frac{L}{2})+(F_B*L)=0\\\\\frac{(W_B*\frac{L}{4}) + (W_A*\frac{L}{2})}{L}=F_B[/tex]

Substitute [tex]\frac{1}{2}W_A[/tex] for [tex]W_B[/tex]

[tex]\frac{W_B}{4}+\frac{W_A}{2}=F_B\\\\\frac{W_A}{8}+\frac{4W_A}{8}=F_B\\\\\frac{5}{8}W_A=F_B\\\\F_B=\frac{5}{8}*1800*9.8=11,025 N[/tex]

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