An engineer wants to move a 100 N force from point A to point B, where B is 0.5 m from A and NOT on the force's line of action. They simply redraw the force acting at point B. What error have they made?
ANone — forces are free vectors and can be relocated anywhere in space without changing their effect
BThey should have moved the force along its line of action to the closest point before relocating it
CBy moving the force off its line of action without adding a compensating couple moment r × F, they have changed the mechanical effect on the body
DForces can only be moved to points that lie on the body itself, not to arbitrary spatial locations
The principle of transmissibility allows a force to slide along its own line of action freely — the mechanical effect is unchanged. But moving a force to a point not on its line of action changes the moment the force produces about every point on the body. To preserve mechanical equivalence, you must add a compensating couple moment equal to r × F, where r is the vector from the new point to the old one. Omitting this couple changes the system.
Question 2 Multiple Choice
You reduce a force system to an equivalent force-couple at reference point A. You then re-express the same system at reference point B. What changes and what stays the same?
ABoth the resultant force and the resultant couple moment change when you shift the reference point
BThe resultant couple moment M_R stays the same; the resultant force R changes based on the new reference
CThe resultant force R stays the same; the resultant couple moment M_R changes to account for the moment of R about the new reference point
DBoth stay the same — equivalent systems are equivalent everywhere, so nothing changes with reference point
The resultant force R is the vector sum of all forces — it doesn't depend on where you evaluate it. But the resultant couple moment M_R depends on the reference point because the moments of all forces must be recalculated about the new point. When you shift from A to B, M_R changes by r_AB × R (the moment of R about the displacement). This is consistent with equivalence: both representations are equivalent to the original system, so they must be equivalent to each other.
Question 3 True / False
Two force systems are mechanically equivalent if they produce the same resultant force, even if their resultant couple moments differ.
TTrue
FFalse
Answer: False
Equivalence requires both conditions: the same resultant force AND the same resultant couple moment about any chosen reference point. The resultant force determines translational effects (ΣF = ma); the resultant couple moment determines rotational effects. Two systems with the same resultant force but different couple moments would cause the same linear acceleration but different angular accelerations — they are not mechanically equivalent.
Question 4 True / False
A force may be freely slid to any other point along its line of action without changing the mechanical effect on a rigid body, but moving it to a point off its line of action requires adding a couple moment to maintain equivalence.
TTrue
FFalse
Answer: True
This is the principle of transmissibility combined with the force-relocation rule. Sliding along the line of action: r × F = 0 (r is parallel to F), so no compensating couple is needed. Moving off the line: r × F ≠ 0, and this couple must be added to preserve the moment effects. The special case (zero couple) is why transmissibility works; the general case requires the compensating couple.
Question 5 Short Answer
Why does the equivalence of two force systems hold everywhere on the rigid body, not just at the reference point where they were shown to be equivalent?
Think about your answer, then reveal below.
Model answer: Because couples are free vectors — a couple moment produces pure rotation with no net force, and its value is the same regardless of where it is evaluated. Once two systems agree on the resultant force R and the resultant couple moment M_R at one reference point, their resultant moments about any other point differ from M_R by exactly r × R — and since both systems have the same R, that adjustment is identical for both. So they remain equal everywhere on the body.
This global equivalence is what makes the reduction useful for structural analysis. A beam cannot distinguish between 100 distributed loads and one equivalent resultant — only R and M_R determine translation and rotation. The abstraction works because equivalence is a global property of the force system, not a local one at the reference point.