Biomechanics For Dummies
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In biomechanics, a common word problem to be solved involves calculating the magnitude of the muscle force required to hold a weight in the hand. A typical problem is worded something like this:

A person holds a 500 Newton (N) dumbbell in his right hand. His forearm and hand are held stationery in the horizontal position with no rotation at the elbow joint. The forearm and hand segment weighs 17 N, and the center of gravity of the forearm/hand segment is 0.23 meters (m) from the axis of the elbow joint. The center of gravity of the dumbbell is 0.34 m from the elbow joint. If the muscle holding the arm in this position inserts 0.05 m from the elbow joint, how much muscle force is required to keep the forearm/hand from rotating at the elbow joint?

Many students are perplexed by how to solve this type of a problem. A step-by-step solution involves first figuring out the biomechanics concept to apply and then selecting and solving the appropriate equation.

"To keep the forearm/hand from rotating at the elbow joint" means to prevent angular acceleration. No angular acceleration is part of a situation called equilibrium. The basic point of equilibrium is that there are no unbalanced forces causing linear acceleration, and there are no unbalanced torques causing angular acceleration of the body. (The body, in this case, is the forearm/hand holding the dumbbell, which is free to rotate at the elbow joint.) So, the biomechanics concept to apply is equilibrium.

Equilibrium, in equation format, is stated as

ΣF = 0 (where F is forces)
ΣT = 0 (where T is torques)

The question describes preventing rotation of the forearm/hand at the elbow joint, which means to maintain ΣT = 0 at the elbow joint. The biomechanics concept to apply is torque.

Torque is the turning effect of a force, calculated as the product of a force (F) and its moment arm (MA), written mathematically as T = F × MA. Before you can sum up the torques, you need to identify the forces that have a moment arm and can create a torque. To do this, go through the problem, identify each force, and give it a label:

  • The weight of the dumbbell can be labeled WD (where D stands for dumbbell).

  • The weight of the forearm/hand segment can be labeled WS (where S stands for segment).

  • The muscle force can be labeled FM (where M stands for muscle).

Weight is a force that always acts downward. Use a plus sign (+) for the upward direction, and a minus sign (–) for the downward direction. Weights are applied at the center of gravity of a body, and the location of the center of gravity for both the segment and the dumbbell weight are given.

Create a table listing what you'll use to calculate the torques, and fill in the known information from the word problem, sort of like this:

Force Magnitude and Direction Moment Arm (MA) Torque (T = F × MA) Torque Name
WD –500 N 0.34 m –170.0 Nm TD
WS –17 N 0.23 m –3.9 Nm TS
FM Unknown, to be solved 0.05 m Unknown TM

It's important that weights be listed as negative forces. The moment arm for each force is on the same side of the elbow joint axis, so set them all as positive. The moment arms for the segment weight and the dumbbell weight are the distance of each center of gravity from the elbow axis because the forearm/hand is in the horizontal position.

The torque created by each force is calculated as the product of the force and moment arm. The weights (segment and dumbbell) create negative torques, and it's important to list the direction, as well as the magnitude of the torque in the table.

Next, use the equation ΣT = 0 to solve for the torque created by the muscle (TM). To do this, expand the equation to list all the torques, like this:

TM + TD + TS = 0

Now, isolate for the unknown muscle torque:

TM = –TDTS

Fill in the known values from the table you created and solve:

TM = –(–170 Nm) – (–3.9 Nm) = 173.9 Nm

The muscle must create a torque of 173.9 Nm, opposite in direction to the torques created by segment and dumbbell weights, to prevent angular acceleration.

The last step is to calculate the muscle force (FM), using the following equation:

TM = FM × MAM

Isolate for FM, making the equation:

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Finally, write out your answer:

The muscle torque required to prevent rotation is 3,478 N. Don't be alarmed when you calculate a large force value from the muscle — the muscle force is always much larger than the force held in the hand, because of the short moment arm for the muscle at the joint.

About This Article

This article is from the book:

About the book author:

Steve McCaw, PhD, is a professor at Illinois State University. Dr. McCaw has taught Biomechanics of Human Movement, Occupational Biomechanics, and Quantitative Biomechanics and has vast experience presenting biomechanics concepts in easy-to-understand formats for use in criminal and civil cases.

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