in an isometric exercise a person places a hand on a scale and pushes vertically
ID: 1979622 • Letter: I
Question
in an isometric exercise a person places a hand on a scale and pushes vertically downward keeping the forearm horizontal. this is possible because the triceps muscle applies an upward force of M perpendicular to the arm. the forearm weighs 22N and has a center of gravity .15m from the elbow. the length of the forearm is .3m> the distance from the elbow to the upper arm bone is .0250m. what is the magnitude of M?cg
.___._________._________________] hand on scale
.0250m .150m
------------.3m---------------------
Explanation / Answer
In an isometric exercise a person places a hand on a scale andpushes vertically downward, keeping the forearm horizontal. This ispossible because the triceps muscle applies an upward force M perpendicular to the arm, as thedrawing indicates. The forearm weighs 25.0N and has a center of gravity as indicated. The scale registers150 N. Determine the magnitude of M. This is more of a statics question than a physics question. We cansimplify the situation so we have a force being exerted down fromthe hand and and a force up near the elbow joint. We know that A) There can be no tendency for the arm to rotate (i.e., no moment- you may know this concept as "torque") at cg, the center ofgravity, since the arm cannot be rotating B) The force exerted down on the scale is equal to the forceregistered by the scale. The moment/torque for a force is Fd, where d is the shortestpossible distance from the line of action of the force to the pointthat the moment is being taken around. Alternatively, theequation is Fdsin? to any line of action. Since the angle forthe shortest line will always be 90 degrees, the equation producesFd. In this case, the lines of action are parallel, so anyperpendicular line will always be the shortest distance. Regardless, we have: - Md1 + (150 N)d2 = 0 (note the minus sign,since the forces are on opposite sides of the pivot point) Simplifying and finding the distances: (150 N)(0.150 m) = M(0.150 m + 0.0250 m) M = 128.6 N
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