Question

The mass of the box is 5kg (the mass center is located in the middle of the box), the mass of a lower arm plus a hand is 1.5 kg (the mass center is located in the middle of the segment), an upper arm weighs 2 kg (the mass center is located in the middle of this segment), and the total mass of the head and torso is 27 kg. Assuming g=10N/kg and 1 iinch=2.54 cm, how large is the force exerted by the back extensor muscles at the L5/S1 in Newtons? How large is the compression force at the L5/S1? How large is the shear force at the L5/S1?

Answer 1

Answer:

Force of Compression= 1,315.322 N

Shear Force = 0 N

Explanation:

Given

Mass of Box = 5kg

Mass of lower arm = 1.5kg

Mass of upper arm = 2kg

Mass of Torso = 27kg

Calculating the weight of the parameters above

Weight of Box = Wbox = 5kg * 10N/kg = 50N

Weight of Lower arm = Wlow = 1.5 * 10 = 15N

Weight of Upper arm = Wup = 2 * 10 = 20N

Weight of Torso = Wtor = 27 * 10 = 270N

The force on the muscle is calculated by the following:

Fmus= Wbox* a/5 + Wlow* b/5 + Wup*c/5 where a,b, and c represents the corresponding lengths of each weights

Calculating a,b and c (See attachment) where angles are 30° and 20°

a = (sin 30) * 14.2 + (cos 20) * 14.5

a = 7.1 + 13.6 = 20.7 in.

b = 13.6 + (7.1/2) = 17.2 in.

c = (14.5/2) cos 20 = 6.8 in

Fmus= 50 * 20.7/5+ 15* 2 * 17.2/5+ 20* 2 *6.8/5 = 364.3N

.... * 2.54 inch

Fmus = 364.3 * 2.54 = 925.322N

Force of compression is calculated as

Fcomp= Fmus + Wtor + Wbox+ 2* Wlow+ 2* Wup

Fcom = 925.322 + 270 + 50 + 2 * 15 + 2 * 20

Fcomp = 1,315.322 N

Fshear = 0 N because no force perpendicular to L5/S1