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 ($\textbf{Fig. E11.11}$). The diving board is of uniform cross section and weighs 280 N. Find (a) the force at the support point and (b) the force at the left-hand end.

Answer

a. $1920 \mathrm{N}$
b. $1140 \mathrm{N}$

Related Courses

Physics 101 Mechanics

University Physics

Chapter 11

Equilibrium and Elasticity

Section 3

Solving Rigid-Body Equilibrium Problems

Discussion

You must be signed in to discuss.

Top Physics 101 Mechanics Educators

Watch More Solved Questions in Chapter 11

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Video Transcript

problem. 11 point. They have someone standing here on a diving board. Um, dimensions are shown. The diving board has a weight of 280 Newtons, which I've already gone ahead, drawn in here halfway along the board Person's weight is 500 Newtons. So what we'd like to do is find the force at the support point here we'll call F one and the Force at the left hand, which will call F too. And we already know the directions of these forces because, you know, there's nothing above the board at this point. So the force has to be coming from below it, and similarly here, uh, I sort of covered it up. But there's just a gap under the board, so it's being supported from the top here, and then force has to be directed down. So the, uh, the, um you know, so we just need to apply our equilibrium conditions And let's call this point the origin. So all of our talks and things are computed about here our day. At least it won't be very helpful for part B because it'll just end up being zero everywhere. That's not so good. So some of the works at this point here, equal zero or about this point, I mean, you know, you add up all of these things being careful about their signs, like the weights of things should be negative and that this is zero meters away from itself. So it ends up going away. You don't have to worry about it s so then you consol ve that for F one and you'll finds this. It'll be the wait Times the distance of the diver, plus the late times the distance to the center of mass of the diving board divided by the distance to the point question one meter from our origin point here. So you concede way have that The Turyk here has to equal the torque downward. Basically, write that out, then divide by distance 2.1 to get a divorce. And so this has a value of 1000 920 Newtons. So that's all good now for For this we can use the fact that the the other equilibrium condition basically some of the vertical forces has people zero Some of the you know the horizontal force is also asked. He's zero, but there's nothing applying any horizontal forces so that you can add Oppa's many zeros as you like and still end up with zero. We don't need to worry about it. So we end up that F two is equal to one minus 280 Newton's from the late of the board, Uh, minus the 500 Newtons of Diver. We know all these things. We just found this. These are given This ends up being 1100 40.

We have video lessons for 97.72% of the questions in this textbook