Statics of rigid bodies: A force applied at a point on the surface of a rigid body may be considered to act at which location without changing the external effect on the body?
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AAt the centre of gravity of the body
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BOn the periphery of the body only
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CAt any point on the line of action of the force
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DAt any point on the surface normal to the line of action of the force
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EOnly at the original point of application
Answer
Correct Answer: At any point on the line of action of the force
Explanation
Introduction / Context:In engineering statics, the “principle of transmissibility” states that a force acting on a rigid body may be shifted anywhere along its line of action without altering the external mechanical effect (resultant force and moment about any point external to the body). This concept is fundamental to simplifying force systems and drawing free-body diagrams.
Given Data / Assumptions:
- The body is rigid (no deformation), so internal stress redistribution is not considered.
- We are evaluating external effects (resultant forces and moments) on the body.
- The same force vector (magnitude and direction) is retained.
Concept / Approach:The external effect of a force on a rigid body is fully described by the force vector and its moment about a reference point. Moving the force along its own line of action does not change the moment about any reference point outside the body, because the perpendicular distance to the line of action remains the same. Thus, equivalent systems result when the point of application is transmitted along the line of action.
Step-by-Step Solution:
Identify the force vector F and its line of action (an infinite line in the force direction passing through the original point).Shift F along this line to a more convenient point for analysis (e.g., to an intersection with a coordinate axis or to a joint).Confirm that the moment of F about any point external to the body remains unchanged when moved along its line of action.Verification / Alternative check:A quick check is to compute the moment M = r × F about a reference O using two different position vectors r1 and r2 to two points on the same line of action; their difference is parallel to F, so (r1 − r2) × F = 0, confirming equal moments.
Why Other Options Are Wrong:
- Centre of gravity: Not required for a general force; only weight acts at the centre of gravity equivalently.
- Periphery or surface-normal points: Arbitrary shifts off the line of action change the induced moment.
- Only original point: Overly restrictive and contradicts the transmissibility principle.
Common Pitfalls:Applying the principle to deformable bodies (where internal effects do change) or shifting a force off its line of action without adding a balancing couple (which would otherwise be needed to retain equivalence).
Final Answer:At any point on the line of action of the force