Use data given. Disregard data in diagram. You can use the following data Body w

Question

Use data given. Disregard data in diagram. You can use the following data Body weight 600N Weight of standing leg 100N Moment arm of the weight of the suitcase = 0.31m Moment arm of the abductor muscle force = 0.05m Moment arm of the body weight 0.05m Moment arm of the body weight excluding the standing leg :0.06m Moment arm of the standing leg = 0.01m (medial to the joint) Moment arm of the ground reaction force 0.06m (medial to the joint) Abductor muscle force (M) acts at a 70 degree angle to the transverse plane (The moment arm is defined as the perpendicular distance from a force) B) Find R and M for the right hip joint when the woman stands on the right leg while carrying a 25kg suitcase in each hand (Fig 7.7b). The moment arm for the weight of the suitcase in her right hand is 0.14m. All other weights and dimensions are as above Which produces a larger hip joint reaction force: carrying 25kg in one hand or 50kg evenly distributed between two hands. Does the distribution of the burden have sufficient effect on the joint reaction force to justify its mention during ergonomic counseling? In other words, does it make more sense to carry twice the load? FMR

Explanation / Answer

These force Produce CW moment about the hip joint

1. Body wieght excluding standing leg = 500*0.05

2. suitcase in left hand                           = 25g*0.31

3. Ground reaction force                         = 600*0.06

Total moment = 136.95 N-m

These   force produce CCW moment

1. standing leg    = 100*0.01

2. right hand suit case   = 25g*0.14

3. Muscle force              = M*0.05

Total moment = 35.3 + 0.05M

equating the CW and CCW moments

35.3 + 0.05 M = 136.95

Muscle force M = 2033 N

Muscle force makes 70 deg with the vertical

FMy = 2033Cos(70) = 695.32 N

FMx = 2033Sin(70)   = 1910.39 N

Joint reaction components FJy , FJx

Now eq. the vertical and horizontal forces

vertical down forces we use -ve and up forces as +ve

1.Body wieght including leg acts vertical down -    -600

2. ground reaction vertically up                          =    +600

3. suit cases    = -50g - down

4. Muscle force FMy   up   1910

FJy -1910 +600 -600 -50*9.8  

FJy = -1420 N

Horizontal forces to the left as -ve and to the right as +ve

In horizontal direction there are only two forces

1. FMx - muscle force

2. FJX - joint reaction

FJx = -FMx = - 1910 N

By carrying two 25 kg in each hand the net moment about the joint reduces than carrying single suit case in one hand. Hence the msucle force is less and the Joint reaction is less.

Related Questions

Navigate

• Browse (All)
• Browse U
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.