Use the following data to calculate numerical values for M,N and B (B will be an

M*sin12*0.7*0.7 = 523*0.42 + 140*0.7......(1) M = 3118.08

ID: 1483124 • Letter: U

Question

Use the following data to calculate numerical values for M,N and B (B will be an angle):

Weight of the torso of the upper body is 523 N and its distance from the lower back is 0.42m

Weight of the box is 140N and its distance from the lower back is 0.70m

The back muscle acts as an angle of 12 degrees above the horizontal at a distance that is roughly 70% of the distamce from the lower back to the arms

Neither the compressive force on the lower back (N) nor it's orientation (B) is known.

I know these three equations:

Sum of Forces acting on X Axis= Nx-Mx=0

Sum of Forces acting on Y Axis= Ny-My-Wt-Ws

Torque= M(Lm sin theta) - (Ws)(Ls) - (Wt)(Lcg)=0

I filled in the torque equation to get:

Torque= M(0.49m sin 12 degrees) - ((140N)(9.8m/s)) (0.70m) - ((523N)(9.8m/s)) (0.42m)

Im not sure if I am on the right track because the value I ended up getting for M seems much to large. I am stuck... please help!

--Also other things I should note, this is a static equilibrium problem

M*sin12*0.7*0.7 = 523*0.42 + 140*0.7......(1)

M = 3118.08 <<-answer

along horizantal Nx = Mx

Nx = 3118.08*cos12

Nx = 3050 N

along vertical

M*sin12 + Ny = 523+140

3118.08*sin12 + Ny = 523+140

Ny = 14.71 N

N =sqrt(Nx^2+Ny^2)

N = sqrt(3050^2+14.71^2)

N = 3050 N

++++

B = tan^-1(Ny/Nx)

M = 3353.98 N

N = -34.33 N

aaangle = 0.28

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