Change the chapter
We analyzed the biceps muscle example with the angle between forearm and upper arm set at $90^\circ$. Using the same numbers as in Example 9.4, find the force exerted by the biceps muscle when the angle is $120^\circ$ and the forearm is in a downward position.
Question Image
<b>Figure 9.28</b> a) The figures shows the forearm of a person holding a book. b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint.
Figure 9.28 a) The figures shows the forearm of a person holding a book. b) Here, you can view an approximately equivalent mechanical system with the pivot at the elbow joint.
Question by OpenStax is licensed under CC BY 4.0.

$471 \textrm{ N}$

Solution Video

OpenStax College Physics for AP® Courses Solution, Chapter 9, Problem 31 (Problems & Exercises) (3:07)

Sign up to view this solution video!


No votes have been submitted yet.

Quiz Mode

Why is this button here? Quiz Mode is a chance to try solving the problem first on your own before viewing the solution. One of the following will probably happen:

  1. You get the answer. Congratulations! It feels good! There might still be more to learn, and you might enjoy comparing your problem solving approach to the best practices demonstrated in the solution video.
  2. You don't get the answer. This is OK! In fact it's awesome, despite the difficult feelings you might have about it. When you don't get the answer, your mind is ready for learning. Think about how much you really want the solution! Your mind will gobble it up when it sees it. Attempting the problem is like trying to assemble the pieces of a puzzle. If you don't get the answer, the gaps in the puzzle are questions that are ready and searching to be filled. This is an active process, where your mind is turned on - learning will happen!
If you wish to show the answer immediately without having to click "Reveal Answer", you may . Quiz Mode is disabled by default, but you can check the Enable Quiz Mode checkbox when editing your profile to re-enable it any time you want. College Physics Answers cares a lot about academic integrity. Quiz Mode is encouragement to use the solutions in a way that is most beneficial for your learning.

Calculator Screenshots

OpenStax College Physics, Chapter 9, Problem 31 (PE) calculator screenshot 1
Video Transcript
This is College Physics Answers with Shaun Dychko. We're going to re-do this question that we've done before in an example but with the angle of between the arm and the bicep going to be 120 degrees. So the hand is down here and the book is on the hand down here. Okay. So, the torques that are going counter-clockwise of which there is only one due to the bicep has to equal the total torque going clockwise. We're assuming the pivot is at the elbow here and so the force exerted on the elbow has no term in our torque formula because its lever arm is zero. We're multiplying each of these forces by the perpendicular component of this distance to the elbow. So, this angle theta which is between the horizontal and the arm is going to be 120 minus 90, and that gives 30 degrees. So the horizontal component of each of these lengths here is going to be the hypotenuse which is the length, multiplied by cosine of thirty. If we consider this triangle for the bicep first of all, it's this triangle here where this is r b and this is r b perpendicular that we want to know, perpendicular to the force. This angle here is 30, that's a right triangle, and so we multiply the hypotenuse by cosine of 30 to get the adjacent. So we substitute that into each of these terms here and we have F b times r b cos theta equals force of weight on the arm, multiplied by the distance from the center of mass of the arm to the elbow, times cos theta, plus force of weight on the book, times the distance from the book to the elbow, multiplied by cos theta. But this is cos theta is a factor on every term and so we can divide both sides by cos theta and it disappears. Then also divide both sides by r b and you solve for the force due to the bicep. So then we plug in numbers. We have two and a half kilograms times 9.81 newtons per kilogram, this is the force of the weight of the arm, multiplied by its 16 centimeter distance to the elbow. I can use centimeters here because since we have centimeters on top and centimeters on the bottom, those units are going to cancel. It doesn't matter what the units are so long as they are the same on top and bottom. You could convert them to meters if you prefer. Then add to this the four kilogram mass of the book times 9.81 times 38 centimeters from the hand to the elbow, divided by four centimeters lever arm of the bicep, or you know, not quite the lever arm technically but it's the distance from where the bicep attaches to the arm bone, the radius or the ulna. Anyway this works out to 471 newtons which is the same as we had in the example by the way. The angle didn't make any difference because it canceled.