Plz help with difficult math

ljlugg

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Mar 10, 2020
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I appreciate any help you can provide for these math problems.

1. A man decides that he wants to build big deltoid shoulder muscles by exercising with dumbbells. He does lateral raises where he keeps his elbow locked and raises his arms horizontally i.e. away from the body. At the peak of the exercise he holds his arm perpendicular to his torso. The dumbbell weighs Wd and his extended arm weighs Wa = 35.0 N and the center of gravity is at the elbow joint. The deltoid muscle is the only muscle acting and the max force the deltoid can apply before it fails is 1880N. This force acts on his humerus at an angle of 15° from the horizontal. 0.150m is the distance between his shoulder joint and his muscle attachment. The distance from his shoulder joint to the center of gravity of his arm is 0.280m and 0.620m is the distance from his shoulder joint to the center of gravity of the dumbbell. The shoulder joint, elbow and center of gravity of the dumbbell are all aligned in the horizontal plane.
a. What is the heaviest dumbbell that he can hold?
b. What are the vertical and horizontal force components that the shoulder joint can apply to
the left end of his arm?

2. Find the force, Fb, that’s exerted by a person’s bicep muscle while they are holding a trophy. The trophy weighs 9.9N and the forearm is parallel to the floor and the elbow is bent at a right angle.
a. Find the force, Fb, that’s exerted by a person’s bicep muscle while they are holding a trophy.

3. This problem involves an ankle joint and the Achilles tendon. A person’s tendon attaches to the calcaneus bone at point P. The Achilles tendon exerts a force, F = 750N, at an angle of 53° horizontal at point P and the distance from the ankle joint to P is 0.037m.
a. What is the moment about the ankle joint? To do this create a right triangle using the ankle joint, P, and ϴ.
 
For question 2. this is what I got. Is this correct?

A force of 9.9 N is acting on a 90-degree angle

T = (9.9N) (90 degrees) (0.01 cm/m) = 8.91
 
If someone can give me the correct formula, I can do the work myself.
 
You (almost) never multiply by the angle itself. You need a trig function of 90 degrees. But sin(90)= 1 and cos(90)= 0. The torque is just force times length of arm.
 
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