Physics A Level
Chapter 4: Forces 4.4 The turning effect of a force
Physics A Level
Chapter 4: Forces 4.4 The turning effect of a force
Forces can make things accelerate. They can do something else as well: they can make an object turn round. We say that they can have a turning effect. Figure 4.16 shows how to use a spanner to turn a nut (a fastener with a threaded hole).
To maximise the turning effect of his force, the operator pulls close to the end of the spanner, as far as possible from the pivot (the centre of the nut) and at ${90^ \circ }$ to the spanner
Moment of a force
The quantity that tells us about the turning effect of a force is its moment. The moment of a force depends on two quantities, the:
- magnitude of the force (the bigger the force, the greater its moment)
- perpendicular distance of the force from the pivot (the further the force acts from the pivot, the greater its moment).
The moment of a force = force $ \times $ perpendicular distance of the pivot from the line of action of the force.
Figure $4.17a$ shows these quantities. The force ${F_1}$ is pushing down on the lever, at a perpendicular distance ${x_1}$ from the pivot. The moment of the force ${F_1}$ about the pivot is then given by:
$\begin{array}{l}
moment\, = \,force\, \times \,dis\tan ce\,from\,pivot\\
= {F_1} \times {x_1}
\end{array}$
The unit of moment is the newton metre (N m). This is a unit that does not have a special name. You can also determine the moment of a force in N cm.
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Figure 4.17b shows a slightly more complicated situation. ${F_2}$ is pushing at an angle $\theta $ to the lever, rather than at ${90^ \circ }$. This makes it have less turning effect. There are two ways to calculate the moment of the force.
Method 1
Draw a perpendicular line from the pivot to the line of the force.
Find the distance ${x_2}$. Calculate the moment of the force, ${F_2} \times {x_2}$. From the right-angled triangle, we can see that:
${x_2} = d\,\sin \theta $
Hence:
$moment\,of\,force = {F_2}\, \times d\,\sin \theta = {F_2}\,d\sin \theta $
Method 2
Calculate the component of ${F_2}$ that is at ${90^ \circ }$ to the lever.
This is ${F_2}\,\sin \theta $. Multiply this by d.
$moment = {F_2}\,\sin \theta \times d$
We get the same result as Method 1:
$moment\,of\,force = {F_2}\,d\,\sin \theta $
Note that any force (such as the component ${F_2}\,\cos \theta $) that passes through the pivot has no turning effect, because the distance from the pivot to the line of the force is zero.
Note also that we can calculate the moment of a force about any point, not just the pivot. However, in solving problems, it is often most convenient to take moments about the pivot as there is often an unknown force acting through the pivot (its contact force on the object).
Balanced or unbalanced?
We can use the idea of the moment of a force to solve two sorts of problem. We can:
- check whether an object will remain balanced or start to rotate
- calculate an unknown force or distance if we know that an object is balanced.
We can use the principle of moments to solve problems. The principle of moments states that, for any object that is in equilibrium, the sum of the clockwise moments about any point provided by the forces acting on the object equals the sum of the anticlockwise moments about that same point.
Note that, for an object to be in equilibrium, we also require that no resultant force acts on it. Worked examples 2, 3 and 4 illustrate how we can use these ideas to determine unknown forces.
WORKED EXAMPLES
2) Is the see-saw shown in Figure 4.18 in equilibrium (balanced), or will it start to rotate?
Figure 4.18: Will these forces make the see-saw rotate, or are their moments balanced?
The see-saw will remain balanced, because the $20 N$ force is twice as far from the pivot as the $40 N$ force.
To prove this, we need to think about each force individually. Which direction is each force trying to turn the see-saw, clockwise or anticlockwise? The $20 N$ force is tending to turn the see-saw anticlockwise, while the $40 N$ force is tending to turn it clockwise.
Step 1: Determine the anticlockwise moment:
$moment\,of\,anticlockwise\,force = 20\, \times 2.0 = 40N\,m$
Step 2: Determine the clockwise moment:
$moment\,of\,clockwise\,force = 40\, \times 1.0 = 40N\,m$
Step 3: We can see that:
clockwise moment = anticlockwise moment
So the see-saw is balanced and therefore does not rotate. The see-saw is in equilibrium.
3) The beam shown in Figure 4.19 is in equilibrium. Determine the force X.
The unknown force X is tending to turn the beam anticlockwise. The other two forces ($10 N$ and $20N$) are tending to turn the beam clockwise. We will start by calculating their moments and adding them together.
Step 1: Determine the clockwise moments:
$\begin{array}{l}
sum\,of\,moments\,of\,clockwise\,force:\, = (10 \times 1.0) + (20 \times 0.5)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,10 + 10\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 20N\,m
\end{array}$
Step 2: Determine the anticlockwise moment:
$moment\,of\,anticlockwise\,force = X\, \times 0.8$
Step 3: Since we know that the beam must be balanced, we can write:
sum of clockwise moments = sum of anticlockwise moments
$\begin{array}{l}
20 = X \times 0.8\\
X = \frac{{20}}{{0.8}}\\
= 25N
\end{array}$
So a force of $25 N$ at a distance of $0.8 m$ from the pivot will keep the beam still and prevent it from rotating (keep it balanced).
4) Figure 4.20 shows the internal structure of a human arm holding an object. The biceps is a muscle attached to one of the bones of the forearm. This muscle provides an upwards force.
An object of weight $50 N$ is held in the hand with the forearm at right angles to the upper arm. Use the principle of moments to determine the muscular force F provided by the biceps, given the following data:
weight of forearm $= 15 N$
distance of biceps from elbow $= 4.0 cm$
distance of centre of gravity of forearm from elbow $= 16 cm$
distance of object in the hand from elbow $= 35 cm$
Step 1: There is a lot of information in this question. It is best to draw a simplified diagram of the forearm that shows all the forces and the relevant distances (Figure 4.21). All distances must be from the pivot, which in this case is the elbow.
Step 2: Determine the clockwise moments:
$\begin{array}{l}
sum\,of\,moments\,of\,clockwise\,forse = (15 + 0.16) + (50 \times 0.35)\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,2.4 + 17.5\\
\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \,19.9\,N\,m
\end{array}$
Step 3: Determine the anticlockwise moment:
moment of anticlockwise force $ = F \times 0.04$
Step 4: Since the arm is in balance, according to the principle of moments we have:
sum of clockwise moments = sum of anticlockwise moments
$19.9 = 0.04F$
$\begin{array}{l}
F = \frac{{19.9}}{{0.04}}\\
= \,497.5\,N\, \approx \,500\,N
\end{array}$
The biceps provides a force of $500 N–a$ force large enough to lift 500 apples!
Questions
7) A wheelbarrow is loaded as shown in Figure 4.22.
a: Calculate the force that the person needs to exert to hold the wheelbarrow’s legs off the ground.
b: Calculate the force exerted by the ground on the legs of the wheelbarrow (taken both together) when the gardener is not holding the handles.
8) A traditional pair of scales uses sliding masses of $10 g$ and $100 g$ to achieve a balance. A diagram of the arrangement is shown in Figure 4.23. The bar itself is supported with its centre of gravity at the pivot.
a: Calculate the value of the mass M, attached at X.
b: State one advantage of this method of measuring mass.
c: Determine the upwards force of the pivot on the bar.
9) Figure 4.24 shows a beam with four forces acting on it.
a: For each force, calculate the moment of the force about point P.
b: State whether each moment is clockwise or anticlockwise.
c: State whether or not the moments of the forces are balanced.
