I do work (6 m)(700 N)/3 = 1.4 kJ of work (plus a bit extra to overcome friction in the pulleys). Well, each of the three sections of rope shortens by 2 m. However, to lift my body through say 2 m in altitude, I must still do (700 N)(2 m) = 1.4 kJ of work. So the force I need supply with my arms is reduced. ![]() So, if we neglect my (modest) acceleration, the force of the three sections pulling upwards on me equals my weight: the magnitude of the tension is T = | W|/3 = (700 N)/3. Compared with g, my acceleration here is negligible. ![]() From Newton's second law, the total force acting on me equals my mass times my acceleration. There are three sections pulling me upwards. The pulleys turn easily, so the tension T in each section of the rope is the same. Here, a single rope goes from the support, down to my harness, round the pulley, back to the support, round another pulley and back to my hands. But not if we use pulleys (which are discussed in more detail in the physics of blocks and pulleys). Lifting my own 70 kg mass (weight W = 700 N)requires more force. Lifting 20 kg bags (weight = 200 N) is not so hard. But the force is in the j (or y) direction and the displacement in the i (or x) direction: they are at right angles. The trolley is supporting six bags, so the force it applies upwards has magnitude 1.2 kN (black arrow), and it moves several metres (red arrow). Look at the big displacement of the trolley and how easy it is. It's possible for a fit person to do a megajoule of work in an hour. The joule is not very big on a human scale: lift a small apple (weight about 1 N) through a height of 1 m and you've done 1 joule of work on the peach – but rather more than that in moving your arm! Similarly, although I only do 140 J on the bag, I do more work on moving my arms and torso. The joule (symbol J) is the SI unit of work: one newton times one metre. So, for the first bag, the force is 200 N, I lift it about 0.7 m (the red arrow), so I do 140 J of work. (which, as we show in the multimedia tutorial, is the increase in potential energy of a mass m in a gravitational field of magnitude g when raised a height Δy). ![]() So, if I apply this constant force over a displacement Δ s = Δx i + Δy j, the work (W) I do is If you remember the scalar product, you'll know that i.j = 0 but j.j = 1. So when I lift them, I'm applying a force F ≅ − W = (200 N) j, which is the black arrow. (Occasionally we shall also need W for the magnitude of W, but you will know from context which is which.) The motion of the bags is slow, their accelerations are small compared to g, so the force required to accelerate them is small compared to their weight. So let's write, for one bag,Ĭheck that notation: weight, W, is a vector, whereas work, W, is a scalar. I did so here to remind you that weight is a vector. Normally we don't say 'down' in this context, because weight is always in a direction close to down. These are 20 kg bags so the weight of each is about 200 newtons down, which is the grey arrow. We begin with the calculations behind the histograms we showed. We all know about physical work, so we started the tutorial with this example, which also gives an idea of the size of the quantities involved.
0 Comments
Leave a Reply. |
AuthorWrite something about yourself. No need to be fancy, just an overview. ArchivesCategories |