Video Transcript
How much torque is produced by a
30-newton force acting on an object at a distance of 0.15 metres from the point
about which the object can rotate?
Now in this question, we haven’t
actually been told what the object in question is. So let’s make life easy for
ourselves and think about it as a simple object, let’s say a ruler. Now for this ruler, if the mass is
evenly distributed along its length, then the centre of mass will be the geometrical
centre of the ruler. And actually this is the point
about which the ruler can rotate. Now, we’ve been told that a
30-newton force is exerted. So let’s say, this is the 30-newton
force. And this force is exerted at
distance of 0.15 metres from the point about which the ruler can rotate.
So here is the point about which
the ruler can rotate, the centre of mass. And here is the point at which
we’re exerting the force. If we assume that the direction in
which we measure this distance, the 0.15-metre distance, is actually perpendicular
to the direction in which the force is applied, then we can recall that the torque
applied on an object is equal to the force exerted on that object multiplied by the
distance between the point at which the force is applied and the point at which the
object can rotate, where the important thing to remember is that the force 𝐹 in
this equation is perpendicular to or at right angles to the distance 𝑑.
And so we can say that the torque
applied 𝑇 is equal to the torque applied which is 30 newtons multiplied by the
distance which we know is 0.15 metres. And once again this works because
we’ve assumed that the force is perpendicular to the distance 𝑑. Then when we evaluate the
right-hand side of the equation, we find that the magnitude of this torque is 4.5
newton metres.
