**Torques in Equilibrium (\(\Sigma \tau = 0; \tau_{cw} = \tau_{ccw}\))**8. A student was given the following problem: "The drawing shows a person (weight, \(W=584\text{N}\)) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position."**A.** To find the answer, the student drew this Free Body Diagram. \(F_h\) is the force on the hands, \(F_f\) is the force on the feet and \(F_s\) is the weight of the man. What is wrong with this diagram?**B.** After the student fixed the Free Body Diagram, they decided to solve the problem by putting the pivot point at the center of mass of the man. Why is this not a good idea?**C.** The student next put the pivot point at the point where the feet touched the ground, but they were concerned because they wanted to calculate the torque: \(F \sin \theta\) but they weren't given any angles and weren't sure what to do. They don't need the angle for this problem, why?**D.** Finally, the student wrote the following equation using the point where the feet touch the ground as the pivot point (they used \(\tau_{cw} = \tau_{ccw}\)):\[F_f (0.840 \text{ m}) = 2F_h (1.250 \text{ m})\]\[F_f \times \frac{584\text{N}(0.840\text{ m})}{2(1.250\text{ m})} = F_h = 196\text{N}\]**E.** Now that we know the force on each hand, find the force on each foot.---The accompanying diagram shows a person doing push-ups with labeled distances and forces. The horizontal distances from the feet to the center of mass (0.840 m) and from the hands to the feet (1.250 m) are shown, indicating where the forces act.

Name: **Torques in Equilibrium (\(\Sigma \tau = 0; \tau_{cw} = \tau_{ccw}\)
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Description: **Torques in Equilibrium (\(\Sigma \tau = 0; \tau_{cw} = \tau_{ccw}\))**8. A student was given the following problem: "The drawing shows a person (weight, \(W=584\text{N}\)) doing push-ups. Find the normal force exerted by the floor on each hand and

Fig: 1 The FBD shown in the picture is wrong because as per the question, it is clearly mentioned that we need to find the force exerted by the floor on each hand and each foot. Hence the direction of Ft and Fh should be reversed as shown in the FBD diagram above(fig 1). Let Ft be the force exerted by the ground on the foot and Fh be the force exerted by the ground on the handChoosing C as a pivot point can be one of the option to solve this problem but it is definitely not a good idea(as per the diagram in the question) Reason: When we are choosing C as the pivot point and calculating sums of moments about the same then we'll have an equation consisting of 2 unknowns, Ft and Fh, for which we need to set a different equation of summation of vertical forces(which is equal to zero) to find the two unknowns. And it will become complex. For this very problem, we don't require any angles since all the forces are vertical and there is no force acting at any angle that needs to resolved into two components namely horizontal and vertical.In this part of the question, the author has taken the foot point as a pivot point( Pt. A as per the diagram in step 1 of the solution) Taking summation of all the moments about point A, MA(total)=0......(Equilibrium)-Fg×0.840+2Fh×(0.840+0.410)=0Fh=584×0.8401.25×2Fh=196.224 N Here, "2Fh" is used because there are two hands and the floor exerts force on both the hands. And, the value 1.25 m is the sum of 0.840 m and 0.410 m, which is used to find the moment for the force FhIn order to find the force on each foot, We take summation of all the vertical forces whose sum is equal to zeroFy(total)=02Ft+2Fh-Fg=0Ft=Fg-2Fh2=584-2×196.2242Ft=95.776 N

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8. A student was given the following problem: "The drawing shows a person (weight push-ups. Find the normal force exerted by the floor on each hand and each foot, as person holds this position." A. To find the answer, the student drew this Free Body Diagram. F, is the force on the hands, F, is the force on the feet and F, is the weight of the man. What is wrong with this diagram? B. After the student fixed the Free Body Diagram, they decided to solve the problem by putting the pivot point at the center of

8. A student was given the following problem: "The drawing shows a person (weight push-ups. Find the normal force exerted by the floor on each hand and each foot, as person holds this position." A. To find the answer, the student drew this Free Body Diagram. F, is the force on the hands, F, is the force on the feet and F, is the weight of the man. What is wrong with this diagram? B. After the student fixed the Free Body Diagram, they decided to solve the problem by putting the pivot point at the center of

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**Torques in Equilibrium (\(\Sigma \tau = 0; \tau_{cw} = \tau_{ccw}\))**

8. A student was given the following problem: "The drawing shows a person (weight, \(W=584\text{N}\)) doing push-ups. Find the normal force exerted by the floor on each hand and each foot, assuming that the person holds this position."

**A.** To find the answer, the student drew this Free Body Diagram. \(F_h\) is the force on the hands, \(F_f\) is the force on the feet and \(F_s\) is the weight of the man. What is wrong with this diagram?

**B.** After the student fixed the Free Body Diagram, they decided to solve the problem by putting the pivot point at the center of mass of the man. Why is this not a good idea?

**C.** The student next put the pivot point at the point where the feet touched the ground, but they were concerned because they wanted to calculate the torque: \(F \sin \theta\) but they weren't given any angles and weren't sure what to do. They don't need the angle for this problem, why?

**D.** Finally, the student wrote the following equation using the point where the feet touch the ground as the pivot point (they used \(\tau_{cw} = \tau_{ccw}\)):

\[
F_f (0.840 \text{ m}) = 2F_h (1.250 \text{ m})
\]

\[
F_f \times \frac{584\text{N}(0.840\text{ m})}{2(1.250\text{ m})} = F_h = 196\text{N}
\]

**E.** Now that we know the force on each hand, find the force on each foot.

---

The accompanying diagram shows a person doing push-ups with labeled distances and forces. The horizontal distances from the feet to the center of mass (0.840 m) and from the hands to the feet (1.250 m) are shown, indicating where the forces act.

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