Define the moment of a force about a pivot.
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the product of the force; and the perpendicular distance from the pivot to the line of action of the force
Moments And Equilibrium
A door handle is 0.65 m from the hinge. A child pushes the handle with a force of 12 N perpendicular to the door. Calculate the moment about the hinge.
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Moment = force × perpendicular distance from the pivot.
The 12 N force is applied perpendicular to the door, so the perpendicular distance from the hinge to the line of action of the force is simply the distance along the door to the handle, 0.65 m.
moment = F × d = 12 N × 0.65 m = 7.8 N m
Moment about the hinge = 7.8 N m
A force of 40 N is applied to the end of a 0.30 m lever at an angle of 35° to the lever. Calculate the moment of the force about the pivot at the other end.
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Finding the perpendicular distance $$x=0.30sin35^{\circ}=0.1721\text{ m}$$ Finding the moment $$M=Fx=40\times 0.1721$$ $$M=6.88\text{ N m}$$
State the principle of moments.
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for a body in rotational equilibrium; the sum of the clockwise moments about a point equals the sum of the anticlockwise moments about the same point
Explain why the centre of gravity of a uniform metre rule lies at its midpoint.
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the rule has uniform density and a symmetrical shape; so the weight is distributed equally about the geometric centre / midpoint
A uniform plank of weight 80 N and length 3.0 m is balanced on a single pivot at its centre. A 50 N child sits 1.2 m from the pivot on one side. Calculate where a 30 N child must sit on the other side to keep the plank balanced.
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Applying the principle of moments $$50\times 1.2=30\times d$$ $$d=\frac{50\times 1.2}{30}$$ $$d=2.0\text{ m from the pivot}$$
A uniform beam of weight 120 N and length 5.0 m rests on two supports, A at the left end and B at the right end. A box of weight 400 N is placed 1.0 m from A. Show that the upward force at support A is 380 N.
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Taking moments about B $${N}_{A}\times 5.0=(400\times 4.0)+(120\times 2.5)$$ $$5.0 {N}_{A}=1600+300=1900$$ $${N}_{A}=\frac{1900}{5.0}=380\text{ N as required}$$
A non-uniform plank of length 4.0 m and weight 200 N rests horizontally on two supports, one at each end. The centre of gravity of the plank is not at its midpoint. When a 150 N child stands at the right end, the upward force at the left support is measured as 90 N. Discuss how this information is used to locate the centre of gravity of the plank. In your answer you should: • calculate the upward force at the right support • use the principle of moments to find the position of the centre of gravity from the left support • explain why the centre of gravity is not at the midpoint.
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Finding the right support force $$\text{total downward force}=200+150=350\text{ N}$$ $$\text{right support force}=350-90=260\text{ N}$$ Taking moments about the left support $$260\times 4.0=(200\times d)+(150\times 4.0)$$ $$1040=200d+600$$ $$200d=440$$ $$d=2.2\text{ m from the left support}$$ The centre of gravity is at 2.2 m, not 2.0 m, so it is not at the geometric midpoint; this means the plank is non-uniform / has more mass on the right half; the principle of moments allows the centre of gravity to be located using only force and distance measurements.