The picture shows a model of a human leg with a heavy weight on the ankle. It re

Question

The picture shows a model of a human leg with a heavy weight on the ankle. It represents a person doing leg-lifts to exercise the quadriceps. In the model, the stick (c.g. in the middle) has a length of 112 cm and a mass of 5.44 kg, while the added mass of 1.90 kg is located 14 cm from the right (non-pivot) end of the stick. The stick makes a 40bdegree angle with the vertical. The stick is supported by a hinge at the left end (pivot) and a rope attached 16 cm from the left end making a 18 degree angle with the stick.

1. What is the magnitude of the gravitational torque (in N*m) due to the weight of the stick?

2. What is the magnitude of the gravitational torque (in N*m) due to the added weight 14 cm from the end of the stick?

3. What is the magnitude of the applied torque (in N*m) due to the rope (muscle)?

4. What is the magnitude of the rope force Fp(in Newtons)?

5. What angle (in degrees) does the rope force make with the vertical?

6. What is the magnitude of the horizontal (x) component of the rope force (in Newtons)?

7. What is the magnitude of the vertical ( y ) component of the rope force (in Newtons)?

8. What is the magnitude of the horizontal (x) component of the joint force (Fj)x(in Newtons)?

9. What is the magnitude of the vertical ( y ) component of the joint force (Fj)y(in Newtons)?

10. What is the magnitude of the joint force Fj(in Newtons)?

11. What angle (in degrees) does the joint force make with the vertical?

Explanation / Answer

1) torque = 5.44*9.8*(1.12/2)*sin(40) = 19.19 N.m

2) torque = 1.9*9.8*(1.12 - 0.14)*sin(40) = 11.73 N.m

3) torque due to rope = 19.19 +11.73 = 30.92 N.m

4)

30.92 = Fp*0.16*sin(18)

Fp = 30.92/(0.16*sin(18)) = 625.4 N

5) 22 defrees


6) Fx = Fp*sin(22) = 234.3 N

7) Fy = Fp*cos(22) = 579.86 N

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