Question

A uniform horizontal lever of length 4 m is pivoted at its midpoint. The following forces are acting on the lever: a 0.63 N vertically upward force acting at the right end of the lever, a 11 N vertically upward force acting at the left end of the lever, and a 9 N vertically downward force acting at a point on the lever 0.5 m away from the right end. Calculate the net torque acting on the lever. (1 point) -52.297 N m -47.275 N m -34.24 N m -63.633 N m -36.76 N m A uniform horizontal lever of length 2.4 m is pivoted at its midpoint. The following forces are acting on the lever: a 7 N vertically downward force acting at the right end of the lever, a 0.63 N downward force (with downward vertical component) that makes an angle of 10 degrees with the horizontal-right acting at the left end of the lever, and a 2 N vertically upward force acting at a point on the lever 0.2 m away from the left end. Calculate the net torque acting on the lever. (1 point) -15.893 N m -13.972 N m -7.852 N m -18.627 N m -10.269 N m A uniform horizontal lever of length 3.63 m is pivoted at its midpoint. A vertically downward force of 3.71 N is acting at the right end of the lever. Calculate the vertically downward force that needs to be applied at a distance of 0.77 m to the left of the pivot if the lever is to be in equilibrium? (1 point) 8.745 N 16.299 N 6.907 N 14.769 N 3.362 N A uniform horizontal lever of length 4 m pivoted at its midpoint is in equilibrium. The following forces are acting on the lever: an 11 N downward force (force with downward vertical component) that makes an angle of 50 degrees with the horizontal-left acting at the right end of the lever, an unknown vertically downward force acting at the left end of the lever, and a 3 N vertically downward force acting at a point on the lever 0.5 m away from the right end. Calculate the magnitude of the unknown force. (1 point) 15.033 N 3.82 N 10.676 N 5.594 N 7.026 N

          A uniform horizontal lever of length 4 m is pivoted at its midpoint. The following forces are acting on the lever: a 0.63 N vertically upward force acting at the right end of the lever, a 11 N vertically upward force acting at the left end of the lever, and a 9 N vertically downward force acting at a point on the lever 0.5 m away from the right end.
Calculate the net torque acting on the lever. (1 point)
-52.297 N m
-47.275 N m
-34.24 N m
-63.633 N m
-36.76 N m

A uniform horizontal lever of length 2.4 m is pivoted at its midpoint. The following forces are acting on the lever: a 7 N vertically downward force acting at the right end of the lever, a 0.63 N downward force (with downward vertical component) that makes an angle of 10 degrees with the horizontal-right acting at the left end of the lever, and a 2 N vertically upward force acting at a point on the lever 0.2 m away from the left end.
Calculate the net torque acting on the lever. (1 point)
-15.893 N m
-13.972 N m
-7.852 N m
-18.627 N m
-10.269 N m

A uniform horizontal lever of length 3.63 m is pivoted at its midpoint. A vertically downward force of 3.71 N is acting at the right end of the lever. Calculate the vertically downward force that needs to be applied at a distance of 0.77 m to the left of the pivot if the lever is to be in equilibrium? (1 point)
8.745 N
16.299 N
6.907 N
14.769 N
3.362 N

A uniform horizontal lever of length 4 m pivoted at its midpoint is in equilibrium. The following forces are acting on the lever: an 11 N downward force (force with downward vertical component) that makes an angle of 50 degrees with the horizontal-left acting at the right end of the lever, an unknown vertically downward force acting at the left end of the lever, and a 3 N vertically downward force acting at a point on the lever 0.5 m away from the right end.
Calculate the magnitude of the unknown force. (1 point)
15.033 N
3.82 N
10.676 N
5.594 N
7.026 N
        
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Added by Joshua H.

University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A uniform horizontal lever of length 4 m is pivoted at its midpoint. The following forces are acting on the lever: a 0.63 N vertically upward force acting at the right end of the lever, a 11 N vertically upward force acting at the left end of the lever, and a 9 N vertically downward force acting at a point on the lever 0.5 m away from the right end. Calculate the net torque acting on the lever. (1 point) -52.297 N m -47.275 N m -34.24 N m -63.633 N m -36.76 N m A uniform horizontal lever of length 2.4 m is pivoted at its midpoint. The following forces are acting on the lever: a 7 N vertically downward force acting at the right end of the lever, a 0.63 N downward force (with downward vertical component) that makes an angle of 10 degrees with the horizontal-right acting at the left end of the lever, and a 2 N vertically upward force acting at a point on the lever 0.2 m away from the left end. Calculate the net torque acting on the lever. (1 point) -15.893 N m -13.972 N m -7.852 N m -18.627 N m -10.269 N m A uniform horizontal lever of length 3.63 m is pivoted at its midpoint. A vertically downward force of 3.71 N is acting at the right end of the lever. Calculate the vertically downward force that needs to be applied at a distance of 0.77 m to the left of the pivot if the lever is to be in equilibrium? (1 point) 8.745 N 16.299 N 6.907 N 14.769 N 3.362 N A uniform horizontal lever of length 4 m pivoted at its midpoint is in equilibrium. The following forces are acting on the lever: an 11 N downward force (force with downward vertical component) that makes an angle of 50 degrees with the horizontal-left acting at the right end of the lever, an unknown vertically downward force acting at the left end of the lever, and a 3 N vertically downward force acting at a point on the lever 0.5 m away from the right end. Calculate the magnitude of the unknown force. (1 point) 15.033 N 3.82 N 10.676 N 5.594 N 7.026 N
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Transcript

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00:01 So in this problem, we're going to be looking at torques, also called moments.
00:05 I'll try to keep it torques like the problem says though.
00:09 So in our first situation here, we have a beam of four meters total, which means each side would be, of course, half of that, or two meters.
00:22 And so we have some forces acting on it.
00:27 Let's not do that in blue, though, because that's the dimension color.
00:37 0 .63 newtons upwards on the right here, 11 newtons upwards on the left here, and then 9 newtons, and this is 0 .5 meters from the end, which means from the center would be 1 .5 meters to match the 2.
00:58 And so for all these problems, we're going to set our axis at the fulcrum here, x and y, and that's where we're going to take our torques about.
01:09 So, of course, we can take the of all twerks and we'll pick counterclockwise to be positive because of course we need a direction.
01:15 So we know that a torque is a force times a distance.
01:20 And so whatever force on the fulcrum doesn't matter because that distance is zero.
01:23 And here we all have perpendicular forces and the direction, so it doesn't really matter to add the angle yet.
01:32 And so if we look at this 11 newton force tends to rotate clockwise around our fulcrum, so it's going to be a negative moment.
01:41 And so the sum of twerks here is a number.
01:43 Equal to negative 11 and what distance is it away from the center? well, it's at the end, so two meters, because that's one half.
01:51 The nine newtons tends to be the same way around.
01:55 So it's also a negative moment 1 .5 meters away from the dimensions we set, and are 0 .63, the other direction, so positive now, also two meters away.
02:06 So we add that all up to get the sum of torques equal to negative 34 .24 and newton meters as the unit.
02:15 That just means our torque is overall clockwise here because we picked counterclockwise to be positive.
02:21 That's the convention, but as long as you're consistent, you can do problems with either direction.
02:25 So the same kind of process for all of our situations.
02:30 Here we have a 2 .4 meter total beam, so it's 1 .2 meters on each side, 7 newtons downwards on one end, and this 0 .63 newton force downwards, but it's towards the left, and it is 10 degrees.
02:48 From the horizontal.
02:50 Now we can break this force down into two components, a vertical force, so we can call it say force in the y and force in the x.
02:59 Now if we look back at our definition of torque, it's based, it's multiplied by a distance.
03:03 So fx here goes along the beam axis and we know our axes are right here.
03:09 And so what that means is the x component of this force makes no moment because its distance is zero.
03:15 So we only want to consider the f y component.
03:17 Now we can also draw it here, we see it's opposite the angle, so it's going to be the sign part of the force.
03:27 So, summa torcs, oh, and there's also a 2 newton force here, which is 0 .2 meters away from this end, so it means it's 1 meter from the fulcrum...
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