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(a) A horizontal force of 20 lb is applied to the handle of a gearshift lever as shown. Find the magnitude of the torque about the pivot point $P$.(b) Find the magnitude of the torque about $P$ if the same force is applied at the elbow $Q$ of the lever.

$\begin{array}{ll}{\text { a) } 40 \mathrm{ft}-\mathrm{lb}} & {\text { b) } 12 \mathrm{ft}-\mathrm{lb}}\end{array}$

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Lily A.

Johns Hopkins University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Boston College

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Video Transcript

Over this question were given that the horizontal force of £20 is applied on the handle of a gearshift lever as shown. Find the manager of the talk about the private point. So this is the private point. B and now we know we know that torque is given by our cross. If where our vector is something which connects the pilot point and the point about which we are applying the force. So this is the art of actor and this is definitely the it factor. So one way is either we find our f scientist to so we find the sine of the angle between these two. Uh that's one way to do it. And uh other way what we can do is uh we can find the perpendicular, perpendicular. Uh The other way is to make the competence of this for so I think finding the angle would be a good choice. So if we expand this, this is the angle which we have to find. So in short this is the angle which we have to find. So that's where this triangle comes. And if we if we connect a parallel line to word and this is let's say the angle tita, then we are looking forward this triangle and in which the opposite side is already given and uh one ft, that is the base is already given. So we can definitely find out sign of angle. Let's talk about the hyper tania's. The hyper tania's and this edge let's call it H one, then h one is going to be square root of one square plus two squared. That's going to be square root of five. So a scientist to is going to be opposite, which is to over hypertension. Such one which is root of five. So it means that the arab actor, the our vector becomes the distance. The R. Factor will be the distance between the force arm and the force arm and point. So we just H. One. So we're gonna write route five Times Force, which is £20 from scientific, which is to over route. So that's going to be 40 40 pound foot. So although other method could have been, we can just have we could just have multiplied just with the Pope in nuclear distance because that's what after all uh matters. So we could have done 20 times too. So we could have got the answer. But anyway, this is also a way uh part B talks about where the manager of the talk about p the same forces applied elbow of at the elbow Q. So I'm going to make a copy of this figure and let's analyze for the part Q. Over here. Now the forces acting over here. So uh What uh magnitude uh of the torque is applied if the same forces applied at you. So this is 20 pounds. So in this case now we're going to use the talk right away. The talk is going to be R cross F. Which means that all we can do is we can multiply the perpendicular distance of R and F between R and F. So the perpendicular distance between this force and this point is 0.6, so we can multiply 0.6 with 20. So this is gonna be 12 £4. This is a required torque in case, too, and this is a required torque in case one. Thank you.

NIT Bhopal

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Vectors

Lily A.

Johns Hopkins University

Heather Z.

Oregon State University

Samuel H.

University of Nottingham

Boston College

Lectures

Join Bootcamp