Answer:
The reactions at the bearings A, B, and C can be found by considering the equilibrium of the rod. The sum of the forces in each direction must be zero, and the sum of the moments about any point must be zero.
The forces acting on the rod are F1, F2, the reactions at the bearings A, B, and C, and the weight of the rod. The weight of the rod can be neglected since it is small compared to the other forces.
The sum of the forces in the x-direction is:
A
x
+B
x
=0
The sum of the forces in the y-direction is:
A
y
+B
y
+C
y
−F
1
=0
The sum of the forces in the z-direction is:
A
z
+B
z
+C
z
−F
2
=0
The sum of the moments about point A is:
B
y
⋅6+C
y
⋅12−F
1
⋅8=0
The sum of the moments about point B is:
A
y
⋅6+C
y
⋅6−F
1
⋅4=0
The sum of the moments about point C is:
A
y
⋅12+B
y
⋅6−F
1
⋅0=0
These equations can be solved to find the reactions at the bearings A, B, and C. The reactions are:
A
x
=0
A
y
=
2
F
1
=400 lb
A
z
=0
B
x
=0
B
y
=
4
F
1
=200 lb
B
z
=0
C
x
=0
C
y
=
4
F
1
=200 lb
C
z
=F
2
=450 lb
Step-by-step explanation: