The problem illustrates a model of a person lifting a 200 N weight. The spine and upper body are represented as a horizontal rod, 70.0 cm long, weighing 350 N, pivoted at the spine base. The erector spinae muscle, attached two-thirds up the spine, stabilizes the back. The spine-muscle angle is 12.0°.Objective: Determine the back muscle tension.**Diagram Details:**- **Person Illustration:** Displays a person bending to lift a weight, showing the pivot point at the base of the back.- **Rod Diagram:** Depicts forces acting on the rod: - The weight of 200 N is at the far end of the rod. - A 350 N force acts downward two-thirds along the rod's length. - Muscle tension (\( \vec{T} \)) acts at a 12.0° angle to the rod. - \( R_y \) and \( R_x \) are components of the reaction force at the pivot.The challenge involves calculating the tension in the back muscle required to maintain this position. **Educational Website Content:**---### Question 1**What is the sum of the torques due to the torso and the weight? (Give your answer in N m.)**- **Answer:** [Input box for answer]---### Question 2**What is the tension in the erector spinalis muscle? (Give your answer in N.)**- **Answer:** [Input box for answer]---### Question 3**This large tension force will compress the spine. What is the force from the base of the spine on the spine? (Give your answer in N.)**- **Note:** The components of this force are labeled as \(R_x\) and \(R_y\) in the figure. To find these, you will need to set both components of the net force equal to zero.**Explanation:** You'll notice that the details of this problem make the y-force almost zero. The x-force is large and leads to a large compression force on the spine. By bending at the waist to pick up a heavy object, the spine has to support a large tension and compression force, which can lead to injury. By bending at the knees to pick up a heavy object, you reduce the risk of back injury!- **Answer:** [Input box for answer]---

We are given weight of spine and its length. We are also given weight which is lifted. We have to find tension in the muscle. We then find reaction force of bottom. We use equilibrium conditions to find the unknown va;ues.Length of spine is L=70cm=0.7m Weight of spine is W=350 N Its distance from left most point is d=L/2=0.35m Weight lifted is F=200 N Its distance from left most point is d'=L=0.7m Hence the total torque produced about bottom is τ=Wd+Fd'τ=350×0.35+200×0.7τ=262.5 Nm Hence the total torque produced about bottom is 262.5 Nm.The tension is T. Component of this perpendicular to length is Ty=Tsin12oTy=0.21T Distance from left end is x=2L/3x=0.47m Torque due to this component must be equal to the total clockwise torque by weight and torso. Hence, the tension is calculated as τ=Tyx262.5=0.21T×0.47T=262.50.21×0.47T=2659.57 N Hence tension is 2659.57 N. Applying equilibrium of forces along horizontal, Rx=Tcos12Rx=2659.57cos12Rx=2601.46 N Applying equilibrium of forces along vertical, 0.21T+Ry=350+200Ry=550-0.21×2659.57Ry=550-552.96Ry=-2.96N Hence the force from base is R=Rx2+Ry2R=2601.462+-2.962R=2601.46 N Hence the force from base is 2601.46 N.

Answered: What is the sum of the torques due to…