A person is doing leg lifts with 3.00-kg ankle weights. The lower leg itself has
ID: 1406926 • Letter: A
Question
A person is doing leg lifts with 3.00-kg ankle weights. The lower leg itself has a mass of 5.00 kg. When the leg is held still at an angle of 30.0 degrees with respect to the horizontal, the patellar tendon pulls on the tibia with a force of 337 N at an angle of 20.0 degrees with respect to the lower leg. Find the magnitude and direction of the force exerted on the tibia by the femur.
The book says that the answer is 281 N, and 39.7 degrees below the horizontal. I need to know HOW this was determined (step by step, as simply as possible please)
Explanation / Answer
force exerted by patellar tendon F1 = 337 N
angle made by the patellar with the horizantal = 20+30 = 50
along horizantal F1x = -337*cos50 = -217 N
along vertrical F1y = 337*sin50 = 258 N
along vertical
gravitational force F2y = (3+5)*9.8 = -78.4 N
Force exerted byfemur = F3
along horizantal = F3x
along vertical = F3y
In equilibrium Fnet = 0
F1x + F2x + F3x = 0
F3x = 217
F1y +F2y + F3y = 0
F3y = -(258-78.4) = -180 N
magnitude = sqrt(217^2+180^2) = 281 N
direction = tan^-1(-180/217) = 39.7 degrees
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