1. Human Arm Equilibrium. A person holds a 1 kg dumbbell at...

1. Human Arm Equilibrium. A person holds a 1 kg dumbbell at rest. Let the forearm have mass 1.06 kg, and assume that the center of mass of the forearm is at point A in the figure. The dimensions shown are rB = 4.5 cm, rA = 14.6 cm, and rD = 38.4 cm. The angle shown is ? = 44 deg. Let us make the simplifying assumption that F acts perpendicular to the line joining O and D. Question: What force F does the bicep exert to maintain this position? Try addressing the question above before reading the following steps. If you get stuck, read on: a. List the equilibrium conditions that apply to the forearm-dumbbell system. Explain how we know (based on fundamental principles applied to the situation of the forearm) that these conditions must hold. b. Draw the free body diagram of the forearm. What is the angle between the rD vector and the FD vector (which is the force that the dumbbell applies to the arm)? c. What is the torque due to FD, computed about the elbow joint O? What makes a torque positive or negative? d. What is the torque due to the force of gravity acting upon the forearm itself, computed about O? e. What torque must force F apply to the forearm to maintain equilibrium? f. What is F?

Transcript

00:01 Hi, here's the scenario.

00:03 A person is holding a 1 .06 kilogram dumbbell and his forearm stays in equilibrium at angle theta with respect to the horizontal axis here.

00:17 Angle theta is 44 degrees.

00:20 The center of mass of the forearm is found at point a where the distance between point a and the elbow joint.

00:30 Which is the pivot where the pivot is.

00:33 You have here it's 14 .6 centimeters from the elbow joint.

00:38 And there's a bicep muscle that is attached to this part here at point b.

00:46 And for a simplified version of this problem, it is said that we will approximate it or consider that the biceps muscle here exerts a nearly 90 degree force with respect to the forearm.

01:07 So if we show the free body diagram of the human arm here now as a beam, then we'd say that, okay, this would be the force of the biceps muscle.

01:19 And the question here is, i mean, the requirement for this problem is to determine the strength of this force of the biceps muscle.

01:27 So it's mentioned that approximately it is 90 degrees with the forearm here.

01:35 Now, after locating the center of mass of the forearm, we know that the total gravitational force of the earth or the weight of the forearm is acting on the center of mass.

01:48 So let's draw the force here due to gravity.

01:52 And we'll just call this gravitational force of the forearm as the weight of the forearm, and that's w subf.

02:02 We have the meaning of the subscripts here, so that's wf.

02:08 And the weight of the dumbbell at the person's hand here also exerts a downward force on the entire forearm.

02:16 So we sketch another force here, and let's call this.

02:21 This is equivalent to the weight of the dumbbell okay now so this is now our free body free body diagram we are to determine the force here of the biceps then we need to use the conditions for equilibrium okay now let's recall first that um for a system to be in rotational equilibrium the total torque that is acting on it at any point it might must be zero.

02:52 And then further recalling the definition of torque, torque is equivalent to the applied force, times the moment arm of the force.

03:02 So the moment arm here is the perpendicular distance between a forces line of action and the pivot.

03:11 So that will help us, that will make our work easier if we focus on, if we do not forget the definition of this point, moment armed.

03:22 It's a perpendicular distance between the pivot and a forces line of action.

03:27 So this gives us the magnitude of the torque of each force.

03:32 And then by sign convention, if the torque causes the object to rotate counterclockwise, then that corresponds to positive torque.

03:42 And clockwise rotation will be corresponding to negative torque.

03:49 Okay.

03:50 So since we know the forces here except for the force of the bicep then our next uh the next thing that we need to do before we add the torque is to determine the individual moment arm or the moment arm of each force okay so let's start with um clarifying with ourselves where the position of the pivot is so it's clear here that the pivot is at the elbow joint so all measurement of moment arm is a reference friends at that point.

04:24 Okay.

04:24 So let's start with the moment arm of the weight of the forearm.

04:30 So as you can see, the weight of the forearm is vertically downward.

04:35 So if you extrapolate for you to see the line of action of w sub f, then you can clearly see that from the pivot going to this distance here, this horizontal distance makes a 90 degree angle, the line of action of wf.

04:55 Hence, this red line here is the moment arm of the weight of the forearm.

05:02 Okay, i hope you can see the right triangle here.

05:06 So if we rewrite it, let's rewrite it somewhere here.

05:11 We will use trigonometry here to determine the moment arm of the weight of the arm.

05:19 So we have here, okay.

05:23 This is theta and then the hypoteness of this triangle is r sub a because it's the distance between the center of mass and the pivot.

05:33 So r sub a, let's write it here as our symbol.

05:37 And we want to determine the base of this triangle, which is the moment arm of the weight of the forearm.

05:44 So i hope you'll agree with me that we will use cosine function to get l sub f.

05:50 So using cosine function, we have lf.

05:54 It will just be the hypotenuse times the cosine of our angle here.

05:59 So r sub a is we need to use the one in meter.

06:03 So we have 0 .146 meter.

06:06 Cosine 44 degrees.

06:09 So that the moment arm, the moment arm of w sub f is, let me look here.

06:18 It is 0 .105 meter.

06:25 Okay, we're done with that.

06:26 Next, let's determine the moment arm of the weight of the dumbbell using the same reasoning.

06:35 So this is the line of action of the weight of the dumbbell.

06:40 So from the pivot, going to this point here, this is obviously making a 90 degree angle with double.

06:49 So this long red line is the moment arm of the weight of the dumbbell.

06:55 So we'll just use the same trigonometric function because the two are similar triangles.

07:04 So we have the moment arm of the weight of the dumbbell...

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