Tom is climbing a 3.0-m-long ladder that leans against a vertical...

Question

Tom is climbing a 3.0-m-long ladder that leans against a vertical wall, contacting the wall 2.5 m above the ground. His weight of 680 N is a vector pointing vertically downward. (Weight is measured in newtons, abbreviated N.) What are the components of Tom's weight parallel and perpendicular to the ladder?

Accepted Answer

Hi everyone in this particular practice problem, we are asked to actually calculate the magnitude of the climber's weight projection along the direction parallel and also perpendicular to the incline part of the wall. So we will have a mountain climber with a weight of 637 Newton using a rope ascending an incline wall and the wall, the wall actually forms a right triangle structure and the wall summit is m above the ground and its base is 4.3 m and is in contact with the ground. We're asked to calculate the magnitude of the climber suite Projection perpendicular and also uh parallel to the wall itself. So I'm gonna start us off with drawing our uh system. So we will have the wall here forming a right angle triangle. So we will have a 90° angle here and we will have the mountain climber being here climbing. Um And I'm gonna draw the perpendicular and the parallel, well, the parallel parallel right now and the perpendicular axis. So this is going to be our um perpendicular axis And this is going to be our parallel axis just like. So and we will have the mountain climber here, having the weight of 637 Newton, uh weight equals 637 Newton. And we were asked, what's the magnitude of the climber's weight projection along this two axis right here? So the weight projection is then going to be along the parallel axis here just like. So I'm just gonna uh bold this a little bit more just in order for you guys to be to make it easier for you guys to see it just like this on the para axis. And this same goes, I'm just gonna indicate this with W of the parallel and the same goes with the perpendicular one. Um It is going to have a projection component just like so pretty similar to the uh parallel one and this is going to be W of the perpendicular. OK. So I am going to call that this angle is going to be the right here. And what we need to know is this data value first And we were given the dimension of the wall, which this length right here is given to be 35 m and the uh length of this bottom part is 4.3 m. So what we can do in order to calculate the angle is to actually use the trigonometry properties. The way we want to do this is to utilize 10 of theta or tangent of theta. We wanna recall that tangent of theta is going to eagles to the opposite over the adjacent. And in this case, this will equals 235 over 4.3 m just like so and in order for us to solve for what the theta is, we want to use the inverse of the tangent here. So the means equals to tangent of to the power of minus 35/4 10.3. I'm just gonna cross out the meters here just like. So or in other words, this is the art tangent of 35/4 0.3, which will result in the theta of seven degrees. So this angle right here is seven degrees. So the weight of the climber W or M G makes the same angle of seven degrees with the incline part of the wall or the, the hypotenuse. So the magnitude of the projection along the direction of either the para or the perpendicular will then be first, I am going to calculate the one for the uh parallel one. So the weight of the parallel will equals to then um M G multiplied by the orientation angle which in this case, since the theta is here, then this angle right here will also then be theta and from there because it is right next to the angle, then this will be cosigned of theta. So this will be the MG or the W is 637 Newton And the cosine of Theta is cosine of 7° and this will result to a valley of 632 Newton. Uh same goes with the uh perpendicular one. This is going to be M G multiplied by because this is uh opposite to denote angle, this is going to be sine theta which will be 637 Newton for weight multiplied by sine of seven degrees. And that will give us a value of Newton. And there, there we go. Uh the magnitude of the climber sued projection along the panel direction is 632 Newton and along the perpendicular direction is 78 Newton which will correspond to option C on our answer choices. So option C here is going to be the answer to this particular practice problem and that will be all for this particular video. If you guys have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.

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