A uniform beam of mass M and length ℓ is mounted on a hinge at...

Question

A uniform beam of mass M and length ℓ is mounted on a hinge at a wall as shown in Fig. 12–101. It is held in a horizontal position by a wire making an angle θ as shown. A mass m is placed on the beam a distance x from the wall, and this distance can be varied. Determine, as a function of x, the tension in the wire.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. A diving board of uniform density has a mass capital M and a length capital L is fixed at one end to the pool deck. It is supported by a wire at the other end which is attached to a point on the deck, making an angle with the horizontal, a diver of mass lowercase M walks to the end of the board which is a distance X the fixed end determine as a function of X, the tension in the wire required to keep the board level. So that's our end goal. Ultimately, we're trying to figure out as a function of X, what the tension in the wire is required to keep the board level. So looking at our diagram here that's provided to us by the prom itself, we have our diver here which is represented by a stick person and he ha and it looks like a he and apparently he has some mass M and he's some distance X away from the wall and it has an angle theta with the board. And the board is denoted by this orange rectangle here. And the length of the board is from the wall, which the wall is this blue bar, this blue rectangle. So from the wall to the end of the bar, like to the end of the diving board is some length L and our diver is some distance X from the wall to where the divers standing. Awesome. We're also given some multiple choice answers. Let's read them off to see what our final answer might be. So A is G divided by cosine of theta. All multiplied by M multiplied by X divided by L plus capital, M divided by two, B is G divided by sin of theta, all multiplied by lowercase M multiplied by X all divided by L plus capital M divided by two. And finally, C is lowercase M multiplied by G all divided by L multiplied by cosine of theta, all of uh and multiplied by X plus capital M multiplied by G all divided by two multiplied by sin of theta. And finally D is M multiplied by G or lowercase M to be exact, M multiplied by G multiplied by L multiplied by sin of theta multiplied by X plus capital M multiplied by G multiplied by sine of theta. OK. Moving right along first off, let us create a diagram of all the forces acting on the diving board. So I've gone ahead and put together a like a free body diagram of sorts of our diving board. And I didn't draw the diver on here to just, you know, save some space and to just focus on the forces here. So as we can see, we have our bolded red arrows to denote our different forces on the far right, we have our vector fa T which is our tension force in the wire. And it has some sort of like the wire makes some angle theta with the board. And the board is denoted by this bolded black line. And our first red arrow going on the bottom going from right to left is we have the vector W which is the weight of the board. And right next to the weight of the board, we have the weight of the diver which we need to note that the weight of the diver is the mass of the diver multiplied by the acceleration due to gravity. And it's the same for the weight of the board. It's the mass of the board multiplied by the acceleration due to gravity. And then from the wall to some distance down the diving board is X and that's where our diver is standing. And then we have some vector F force of the wall. So we have the vector f force of the wall that's acting on the board itself. And from the wall to the board or where the board ends is some length divided by two. Awesome. So now that we have a visualization of what's going on here, we need to note that a force due to the diver's weight can be written as and I discussed this verbally, but I can write this out, you know, too. So the weight of the diver, which I'm gonna call W subscript D is equal to the mass which is lowercase M multiplied by G which is the acceleration of due to gravity. And this will act in the downwards direction at a distance X from the fixed end, a force due to the board's weight can be written as WB is equal to capital M multiplied by G where little M is the mass for diver once again and capital M is the mass of the diving board. And this one WB will act in the downwards direction at the board center of mass which is L divided by two from the fixed end due to the fact that the board has uniform density. So a 4 to 10 which we can write as F subscript capital T acts at the end of the board. L from the fixed end, this force will make an angle theta with the horizontal but only the horizontal, sorry only the vertical component. So the vertical component only which let's write this down. I will make a little note here like a little star here. So VC I'm gonna call it V dot C. So this is the vertical component is F T multiplied by S and this is using Trigon trigone trigonometry uh using trigonometry. So trig principles to figure out. So sign of theta and this contributes to the torque about the pivot point which is the fixed end and the force exerted by the wall on the board contributes no torque about the pivot which is the fixed end on the left. So now we need to calculate the tension force ft. So in order to do that, we need to start by considering the torque forces acting on the board, let us note that the net torque is zero. Since the board is in equilibrium, let us also consider that the torque would cause a counterclockwise rotation to be negative and the one that causes the clockwise rotation to be positive. Therefore, we can write this particular equation as and we'll use Tau to denote Tau, which is the Greek symbol Tau to denote torque. And we'll call it the torque and the attention force. So torque T minus FT multiplied by sine of theta multiplied by capital L is all equal to zero. Now, we need to rearrange this equation to isolate and solve for Tau T. Tau T is equal to FT multiplied by sine of seda multiplied by capital L which we can write that the torque on the diver plus the torque on the board awesome is all equal to the force of the tension multiplied by sine of theta multiplied by capital L which we can also write as lowercase M multiplied by G multiplied by X plus capital M multiplied by G multiplied by capital L all divided by two, which is all equal to 10 force multiplied by sine of theta multiplied by capital L. So now we can rearrange this equation to isolate and sulfur FT which will be our final answer. So FT is equal to M or I should say lowercase M multiplied by G multiplied by X plus capital M multiplied by G multiplied by capital L all divided by two. And this is all divided by L multiplied by sine of theta. And this is all equal to. Now we're just simplifying down to as we're simplifying as many times as it takes to get it into its simplest form. So lowercase M multiplied by G all divided by L multiplied by sine of theta. And this is multiplied by X plus capital M multiplied by G divided by two, multiplied by sine of theta. Awesome. So sign of the, which all of this is equal to, we could simplify one last time and this is all equal to G divided by sign of theta multiplied by lowercase M multiplied by X divided by capital L plus capital M divided by two. And that's it we've solved for this problem. Hooray, we did it. So looking at our multiple choice answers the correct answer has to be the letter B which once again, it's G divided by sin of theta all multiplied by lowercase M multiplied by X all divided by L plus capital M divided by two. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Torque

Equilibrium

Trigonometric Functions

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