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 components of the force exerted by the beam on the hinge.

Diagram of a beam hinged to a wall, with a mass on it and a wire at an angle, illustrating rotational equilibrium concepts.

Accepted Answer

Hi, everyone. Let's take a look at this practice problem dealing with static equilibrium. In this problem, we have a uniform horizontal signpost of mass capital M and length L and it's fixed to the vertical pole on the left, a support cable on the right which makes an angle theta with respect to the horizontal um stabilizes a signpost. And there is also a bird that lands on this post and it, the bird has a mass little M and it's positioned to distance X from the pole. Question was to determine the vertical and horizontal components that the po uh horizontal signpost exerts on the pole at the point where it's attached. Below. The question, we're given a diagram of what was described in the problem. We're also given four choices. Each choice um gives an expression for the vertical and horizontal um components of the force that we're looking for in terms of the two masses, the distance X the length L and the angle theta. Now, the first thing we want to do is draw a free body diagram or our um horizontal signpost. So I'm gonna draw a rectangle here and on this rectangle, I am going to um label each of the forces acting on um the signpost. On the right hand side of this post, we have the tension force will label as FT and it makes an angle theta with respect to the horizontal. We're also gonna have the weight of the signpost itself. And we're told that we have a uniform signpost. So the weight is going to act at the center of gravity and for a uniform object center of gravity is um at the uh geometric center. And so at the center of the post, we'll have the weight going down and the weight is gonna be a capital MG. And it's gonna be located a distance L over two from the left end. The next force is gonna be the weight of the bird. So we're gonna have um going downwards a little MG and that's gonna be located a distance X from the left end of the beam. And finally, we'll have the unknown force that we're looking for and we have no idea what direction it's going to be going. So I'm just going to have it going up into the right label it as F and we're actually gonna break this into components. We'll have F horizontal in the X direction and we'll have F vertical in the Y direction and in terms of um directions, we're gonna add a coordinate system here. So we'll have the positive X direction going to the right and the positive Y direction going up. Now we're dealing with stack equilibrium. So we actually have two conditions. We need to worry about, we need to have the net torque equal to zero and the net force equal to zero. So our net Twerk. So if I thumb the torques, let's give them a net torque and that's gonna have to be equal to zero. And just to remind you um our torque, we're looking for the magnitude of that torque it's going to be equal to RF time data where R is the distance from the pivot point to the location of the force F is the force and the angle theta is the angle between the R and the F vectors. Now, since we're dealing in uh with two dimensions here, um our net force equation can be broken up into components. So I'll have the um net force in the X direction as to equal zero. And the net force in the Y direction is also going to have to equal zero. So those are the equations that we're going to use to set up the problem. Let's start with the net torque equation. So our network in order to use this um net torque, um we need to pick a pivot point. Now we're free to choose any pivot point that we want to. Since we're done with stack equilibrium, we should have zero at every single pivot point. If not, the object would begin to rotate at that point. Um And in problems like this, it's advantageous to put our um pivot point at the location where we have the most unknowns. And so that's actually going to be here on the left end of our um post because I don't know the vertical, horizontal components. And so by doing so when I look at the torque by those two horses, they're gonna actually equal zero because I'll have zero times whatever those forces have to be. Since R in that case is gonna be zero. So for my net torque, my first term is actually going to be zero. I then need to um look at the other three forces and calculate their torques. So for the weight of the bird, um I'm going to first have to determine the direction of the torque, a torque is a vector quantity and we get its direction by using the right hand rule. So I to get the direction of the torque you point your fingers in the direction of R which in this case is going to the right, so it's going from the pivot point to the um force, I curl my fingers in the direction of the force. This case is downward and my thumb points in the direction of the torque which is going into the screen. So that's gonna be in the negative Z direction. So this torque is actually gonna come in with a negative sign. So I'll have minus, I'll have the distance, the torque is acting from the pivot point that's gonna be X, the force which is little NG. And then I'm gonna have um sign of the angle between those two vectors which in this case is 90 degrees. So I have um multiply that by the sine of 90 degrees, however, s 90 degrees is one. So I'm not going to write that term down. We'll do the same thing for the um weight of the beam. It's also gonna come in with a negative sign. It's gonna be a distance L over two from the pivot point. Um The force is gonna be capital MG and we're also gonna multiply it by sign of 90 degrees, which is one now for the um torque due to the tension force, um that's gonna be a positive torque. We're gonna have a plus, its distance is going to be L from pivot point since it is at the end of the beam. Um I don't know what the force is. So I'll have ft as the force and then I will have to multiply this by the sign of data because that's gonna be the angle between the R and the F vectors. And then finally, I can set that all equal to zero. Now, at this point, the only unknown in this um equation here is the tension and I'm not asked to actually find the tension. I'm asked to find the vertical hor the components of the force. However, when I do the net forces in both the X and the Y direction, the tension force will show up. So it's a good idea to go ahead and solve for it here. So I can plug it in later. So I can actually solve this for the tension force. So I have FT is equal to X little MD plus L divided by two capital MG, that whole quantity divided by L stein theta. So now we can move on to the uh net force in the X direction. All some of the forces in the X direction. There are actually two of them. I have the horizontal force which is what I'm looking for and that's going in the positive X direction. Then I'll have a component of the tension going in the negative X direction. So I'll have minus FT and to get the um X component, I wanna multiply that by cosine data and all that has to equal zero. So, so for the horizontal um force, I have F horizontal is equal to and in place of the tension, I'm gonna write down the expression that we had found. So we have X little MG plus L over two capital MG, all divided by I find the and that whole fraction is multiplied by cosine data. I can actually simplify this a little bit. So I have F horizontal, I want to break that um one fraction into two smaller fraction, they'll have little MG divided by elf. And I can actually combine the cosine and sine here. So I have cosine divided by sine and that gives me one over tangent. So I'll put this tangent here, tangent of theta in the denominator. And that whole fraction is still being multiplied by X. They all have plus capital MG divided by two. The LS would cancel and I'm still left with a tangent of data and the denominator. So that is the first half of our answer for our um what we're looking for though the F horizontal is equal to um little MG divided by L tangent. The all multiplied by X plus capital NG divided by two tangent data. So now we can um look the vertical component using the thumb of the forces in the Y direction, some of the forces in the Y direction. I actually have four of these forces. I have the vertical component of the force, which is what I'm looking for going in the positive Y direction minus the weight of the bird which is minus little MG minus capital MG, which is the weight of the post. And then I'll have plus the vertical component of the tension, the plus ft um sine data. So the sine data there um gives me the vertical component and all of that has to equal zero. So we can solve now for the vertical component of the worse the F Vert vertical is equal to little mg plus capital mg minus. And in place of the tension will have our expression that we found earlier. There'll be X little mg plus L over two capital mg all divided by L sine theta. And that whole fraction gets multiplied by sign the, and we can actually simplify this a little bit. First thing we can do, um, the sign is we'll cancel out and I can collect um like terms here. So I have little MG and if I click terms that have low MG in them, I'll have low MG multiplied by the quantity of one minus X divided by L. And then I'll have plus um Now I can look for terms with capital MG in them. So I have capital MG minus. Um The LS will cancel and so I'm left with minus one half M capital MG that leaves me overall with a positive plus one half capital MG. And that's actually the answer to second half of our answer or the vertical component. So I can now look at these um two expressions here and look at our solutions. And the only solution that matches up with that is actually b the other solutions either have wrong factors. I'm with them or the wrong trick functions. So just sort of recap, we've used our um conditions for static equilibrium that the net torque and the net force has to be equal to zero. And we use that to solve for our unknown intention and the um vertical and horizontal components of the horses and um through these three equations. So I hope that this has been helpful and I'll see you in the next video.

Key Concepts

Torque

Equilibrium

Force Components

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