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 perpendicular to the arm, as the drawing indicates. The forearm weighs 25.0 N and has a center of gravity as indicated. The scale registers 85 N. Determine the magnitude of M.
The forearm weighs 25.0 N and has a center of gravity as indicated.
We do not see the diagram, and there is no description. We do not see what M represents.
In any case, sum moments about the support point of the muscle, at which the reaction is not known. The sum of moments should be zero. Solve for M.
To determine the magnitude of M, we need to analyze the forces acting on the system and apply Newton's laws of motion.
Let's break down the forces involved:
1. Weight of the forearm: The forearm weighs 25.0 N acting straight down at its center of gravity.
2. Force exerted by the triceps muscle: The triceps muscle applies an upward force perpendicular to the forearm. Let's call this force Ft.
3. Force registered by the scale: The scale registers a force of 85 N, which is the sum of the weight of the forearm and the force exerted by the triceps muscle.
Now, according to Newton's laws of motion, the net force acting in the vertical direction must be zero since the forearm is not accelerating vertically.
So, let's write the equation for the net force:
Net force = Force exerted by the triceps muscle - Weight of the forearm
Since the net force is zero, the two forces must be equal:
Ft - 25.0 N = 0
Simplifying the equation, we find:
Ft = 25.0 N
Therefore, the magnitude of M is 25.0 N.