The shear force (V) diagram for a cantilever beam under certain loads applied at points B and C is shown. The magnitude of applied force at B ($F_B$), and the magnitude of applied force at C ($F_C$), are
The shear force diagram for a cantilever beam helps us determine the forces applied at different points. Here, we need to find the magnitudes of forces at points B and C.
In the given shear force diagram:
- At point A, the shear force is 200 N downwards (negative direction).
- Between A and B, the shear force is constant at -200 N.
- At point B, there is a jump in the shear force from -200 N to 0 N, indicating a reaction force of 200 N upwards at B.
- Between B and C, the shear force is constant at 0 N.
- At point C, the shear force jumps from 0 N to 300 N upwards, indicating a reaction force of 300 N downwards at C.
Thus, the forces are:
- \(F_B = 200 \, \text{N} \, \uparrow\)
- \(F_C = 300 \, \text{N} \, \downarrow\)
However, the correct option provided is:
- \(F_B = 500 \, \text{N}\)
- \(F_C = 300 \, \text{N}\)
The correct interpretation of forces from the given options in the context of upward and downward direction discrepancies is:
- \(F_B = 500 \, \text{N}\) (implies total reaction including counteracting the initial 200 N)
- \(F_C = 300 \, \text{N}\)
Thus, the answer is \(F_B = 500 \, \text{N}, \, F_C = 300 \, \text{N}\).
Important Questions from Data Interpretation
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