A diving board 3.00 m long is supported at a point 1.00 m from the end, and a diver weighing 500 N stands at the free end (Fig. E11.11). The diving board is of uniform cross section and weighs 280 N.

Find (a) the force at the support point.

Video transcript

Hey everybody today, we're dealing with a problem about equilibrium. So we're being told that a base athlete, base jumper athlete carrying a parachute stands at the free end of a four m long homogeneous board that is placed at the top of a skyscraper. Now the board is fixed on the left end, is fixed on the left hand and lies on the support P. That lies at a distance d of 1.5 m from the left hand end where it's supported and fixed. If the jumper weighs 650 newtons With the parachute and the board itself weighs 300 newtons were being asked to determine what is the force at the support point? In other words, what is the force exerted by the by the support point on the board itself? So there's a few things to take note of here. one. This system is uh at rest, right. Which means that the second condition of equilibrium that states that the sum of all torque within the system Equal to zero. That second condition is fulfilled because it's fulfilled. We still need to take. We know that the sum of all torques must be zero. But what is the actual some of the torques? What are the individual values that we need to take in mind here and to do that. We need to take the some of the talks about the left point where it's being supported. Now, let's make this a sort of graph, right, have an X. Axis and we have a why axis. There is a force being exerted by the support point f. one. This is what we're trying to find. There's also a force being exerted down at the point where it's actually fixed F two. However, that's being acted upon by a uh well it's fixed, so we don't really count that here as well as the fact that there's a normal force pointing the opposite direction, that sort of cancels that out force. Other than that we have a couple other, we'll not force but parameters taking to count. Well, it is a force, I suppose we have the weight of the jumper, 600 newtons as well as the weight as well as the weight of the board. 300. So the weight of the jumper again, 650 Newtons. But with this in mind we can go ahead and start actually finding the some of the torques right? We know it's equal to zero, but what is the actual expansion of that? So let's let's do that. The torques is equal to zero To go to F one times The 1.5 m minus the weight of the board. one of the board ups, That's because it's 1.5 m to the support point where the weight of the board. We see As well as multiplied by 2.0 m minus the weight of the jumper, multiplied by four m And it's four m because that is the whole length of the board. Now. Again, as I mentioned earlier, we're trying to solve for F1 and we know that this will also be equal to zero. So rearranging and solving for F1, the force on the support point, we get 300 Newtons Times two m plus 615 Newtons times four m, Divided by 1.5 m, which gives us a final answer Of Newtons or Answice Choice D. This means that the force at the support point, given all the parameters that we have here will be 2130 Newtons. I hope this helps. And I look forward to seeing you all in the next one.