Turning Effect Of Forces

The magnitude of the force; the perpendicular distance from the pivot to the line of action of the force.

The moment of a force equals force x perpendicular distance from the pivot. Placing the handle far from the hinge increases the perpendicular distance; therefore, a smaller force is needed to produce the same turning effect / moment to open the door.

The moment of a force is the product of a force and the perpendicular distance from the pivot.

Finding the moment Equation used $$M=F\times d$$ Given $$F=12\text{ N}$$ $$d=0.40\text{ m}$$ Substituting: $$M=12\times 0.40$$ $$M=4.8\text{ N m}$$ The moment is 4.8 N m. Moment is directly proportional to perpendicular distance when force is constant; therefore, if the perpendicular distance doubles (from 0.40 m to 0.80 m), the moment also doubles, from 4.8 N m to 9.6 N m.

For an object in equilibrium, the sum of the clockwise moments about any pivot equals the sum of the anticlockwise moments about that pivot.

Finding the unknown weight Applying the principle of moments: $$\text{anticlockwise moment}=\text{clockwise moment}$$ $${F}_{1}\times {d}_{1}={F}_{2}\times {d}_{2}$$ Rearranging for ${F}_{2}$: $${F}_{2}=\frac{{F}_{1}\times {d}_{1}}{{d}_{2}}$$ Given $${F}_{1}=4.0\text{ N}$$ $${d}_{1}=0.60\text{ m}$$ $${d}_{2}=0.80\text{ m}$$ Substituting: $${F}_{2}=\frac{4.0\times 0.60}{0.80}$$ $${F}_{2}=\frac{2.4}{0.80}$$ $${F}_{2}=3.0\text{ N}$$ The required weight is 3.0 N.

There is no resultant force acting on the object; there is no resultant moment acting on the object.

The beam is in equilibrium because the upward forces from the two supports equal the downward weight of the beam, so there is no resultant force. The sum of the clockwise moments about any point on the beam equals the sum of the anticlockwise moments about that point, so there is no resultant moment. Both conditions for equilibrium are satisfied; therefore, the beam does not accelerate or rotate.

Finding the total clockwise moment $${\text{Moment}}_{1}=2.0\times 0.30=0.60\text{ N m}$$ $${\text{Moment}}_{2}=4.0\times 0.10=0.40\text{ N m}$$ $$\text{Total clockwise moment}=0.60+0.40=1.0\text{ N m}$$ The total clockwise moment about the pivot is 1.0 N m.

When an object is in equilibrium, the sum of the clockwise moments equals the sum of the anticlockwise moments about any point; this is because both the resultant force and the resultant moment are zero. Choosing a different point changes the individual moment values, but the clockwise and anticlockwise totals still balance; therefore, the principle holds regardless of the chosen point.

The student should adjust the pivot position until the beam balances horizontally with no loads; this ensures the pivot is at the centre of mass of the beam, so the weight of the beam itself does not produce a resultant moment. If this is not done, the beam's own weight creates an unaccounted moment that causes errors in the results.

Finding the required weight Applying the principle of moments: $$\text{sum of anticlockwise moments}=\text{sum of clockwise moments}$$ Calculating anticlockwise moments (left side): $${\text{Moment}}_{1}=3.0\times 0.30=0.90\text{ N m}$$ $${\text{Moment}}_{2}=1.0\times 0.10=0.10\text{ N m}$$ $$\text{Total anticlockwise}=0.90+0.10=1.0\text{ N m}$$ Setting equal to the clockwise moment: $$1.0=W\times 0.20$$ Rearranging for $W$: $$W=\frac{1.0}{0.20}$$ $$W=5.0\text{ N}$$ The weight required for balance is 5.0 N.