A 70.5 kg weight lifter is doing arm raises using a 7.50 kg...

A 70.5 kg weight lifter is doing arm raises using a 7.50 kg weight in her hand. Her arm pivots around the elbow joint, starting 40.0 ° below the horizontal. (See the figure(Figure 1).) Biometric measurements have shown that both forearms and the hands together account for 5.90 % of a person's weight. Since the upper arm is held vertically, the biceps muscle always acts vertically and is attached to the bones of the forearm 5.50 cm from the elbow joint. The center of mass of this person's forearm-hand combination is 14.0 cm from the elbow joint, along the bones of the forearm, and the weight is held 38.0 cm from the elbow joint. Part A What force does the biceps muscle exert on the forearm? F = 513.11 N Part B Find the magnitude of the force that the elbow joint exerts on the forearm. Part C Find the direction of the force that the elbow joint exerts on the forearm. upward downward

Transcript

00:01 So in this problem we have our weight lifter and our free body diagram is going to be the lower part of the arm.

00:08 So our axis we can pick right at where the elbow connects and we're going to have x and y, x horizontal not along the beam as we usually do because all of our forces are in the vertical direction and so we want one of our axes vertical.

00:25 Now we have the weight at the end of the arm 7 .5 kilograms multiplied by g, the exertion to gravity to get weight, and the weight of the arm itself.

00:40 Now both arms are 6 .3 percent, both arms and hands, since we just have one arm in hand we want half of that or 3 .15 percent.

00:57 So the weight of the arm or the mass of the arm is the 71 kilograms that the entire person weighs times 0 .0315 or 2 .252 kilograms.

01:10 So that is the weight 2 .5 or 2 .252 times g to get the weight.

01:21 Now there is a vertical tension from the force of the biceps or fb, i didn't feel like subscript so it's a tension force here, and a force from the elbow.

01:32 Now i already know it's going to be in the downward direction so we'll write it.

01:35 If you don't know which direction it is just assume in the positive you get a different sign you know you've just picked the wrong direction which is okay.

01:41 And of course that's the force at the elbow.

01:46 So we need dimensions here and measuring everything along the length of the arm which is what it seems like, not 100 percent sure on this but what the rest of the problems shows that's what it seems like, 5 .5 centimeters or 0 .055 meters along here.

02:08 Then for the mass of the arm which is specified 17 centimeters 0 .17 meters and the total length of the arm 0 .38 meters 38 centimeters.

02:25 And we see where we're going with this we're going to look at sum of moments.

02:28 When equilibrium so the sum of moments is equal to zero counterclockwise positive by convention.

02:37 So the tension force is a positive moment based on taking moments about our center of our origin here.

02:46 Tension is positive, the weights have a negative moment they're rotating the other direction, and the elbow force makes no moment because it's right on the origin.

02:58 And we need to know our angle for this because we need the perpendicular distance so we want the component of distance here and for all these we want the adjacent to the angle component because we have the hypotenuse so we're going to use the cosine 40 for everything.

03:18 So starting with the tension that's our force and then the distance 0 .055 cosine 40.

03:31 And then for the mass of the arm 2 .252 9 .81 for the acceleration due to gravity and then distance 0 .17 cosine 40.

03:45 And same exact way for the weight at the end of the arm, mass, gravity, and distance.

03:53 And that's all equal to zero.

03:58 Now we notice we have a cosine 40 in every term and so we can factor that out.

04:03 So we see because we've measured everything along the beam the actual angle of the beam won't matter which we'll get to a bit later...

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