**Physics Problem: Muscle Tension and Force on the Spine****Part A****Question**: What is the tension in the erector muscle? *Hint:* Align your x-axis with the axis of the spine.**Instructions**: Express your answer with the appropriate units.- **Input Box**: - Variables/Icons: Calculator, micro symbol, question mark for help - Placeholder Text: "Value" "Units"**Feedback**: - Incorrect; Try Again; 5 attempts remaining---**Part B****Question**: A force from the pelvic girdle acts on the base of the spine. What is the component of this force in the direction of the spine? (This large force is the cause of many back injuries.)**Instructions**: Express your answer with the appropriate units.- **Input Box**: - Variables/Icons: Calculator, micro symbol, question mark for help - Placeholder Text: "Value" "Units"**Action Button**: - Submit- Request Answer When you bend over, a series of large muscles, the erector spinae, pull on your spine to hold you up. Figure 1 shows a simplified model of the spine as a rod of length \( L \) that pivots at its lower end. In this model, the center of gravity of the 320 N weight of the upper torso is at the center of the spine. The 160 N weight of the head and arms acts at the top of the spine. The erector spinae muscles are modeled as a single muscle that acts at a 12° angle to the spine. Suppose the person in Figure 1 bends over to an angle of 30° from the horizontal.**Diagram Explanation:**- The diagram illustrates a person bent over at an angle of 30° from the horizontal.- The spine is represented as a rod of length \( L \), pivoting at the lower end.- There are two points of interest: - The center of gravity of the upper torso, located at the halfway point of the rod, marked \( L/2 \). - The center of gravity of the head and arms, located at the top of the rod.- An arrow represents the force exerted by the erector spinae muscles, labeled "Effective location of erector spinae muscles," acting at a 12° angle to the spine.

Given:- A simplified model of the spine as a rod of length L that pivots at its lower end. the center of gravity of the 320 N weight of the upper torso is at the center of the spine. The 160 N weight of the head and arms acts at the top of the spine. The erector spinae muscles are modeled as a single muscle that acts at a 12° angle to the spine. The person bends over to an angle of 30° from the horizontal. Find:- What is the tension in the erector muscle? Hint: Align your 3-axis with the axis of the spine. A force from the pelvic girdle acts on the base of the spine. What is the component of this force in the direction of the spine? Solution:- a) We first make a free body diagram of the system as Now We know that in the equilibrium condition, the net torque acting on the system is Zero. ∑τnet= 0Now Wh×Lcosθ+Wt×Lcosθ2-T×2L3sinϕT×2L3sinϕ=Wh×Lcosθ+Wt×Lcosθ2T= Wh×Lcosθ+Wt×Lcosθ22L3sinϕT= Wh×Lcosθ+Wt×Lcosθ22L3sinϕ Now put the given value in the equation of T T= Whcosθ+Wt×cosθ223sinϕT= 170cos30+360×cos30223sin12T= 2186.809≈2186.81 N T=2186.81 N The tension in the erector muscle is 2186.81 N b) We know that in the equilibrium condition the first law, The force pelvic girdle acts on the base of the spine is F = Wtsinθ+Whsinθ+TcosϕF= 360sin30+170sin30+2186.81cos12F= 2404.023 N A force from the pelvic girdle that acts on the base of the spine is 2404.023 N