Attempt 3 of 3G Chapter 12, Problem 076 Your answer is partially correct. A gymn

Question

Attempt 3 of 3G Chapter 12, Problem 076 Your answer is partially correct. A gymnast with mass 42.0 kg stands on the end of a uniform balance beam as shown in the figure. The beam is 5.20 m long and has a mass of 270 kg (excluding the mass of the two supports). Each support is 0.540 m from its end of the beam. In unit-vector notation, what are the forces on the beam due to (a) support 1 and (b) support 2 (a) Number || 9019.32 (b) Number l 9019.32 Units N LINK TO TEXT SCORE 1.0 Maximum Point Potential:

Explanation / Answer

(a) For computing torques, we choose the axis to be at support 2 and consider torques that encourage counterclockwise rotation to be positive. Let m = mass of gymnast and M = mass of beam. Thus, equilibrium of torques leads to

Mg(2.60m) mg(0.54m) F1(5.20m) = 0.

F1 = (270*9.81*2.6 - 42*9.81*0.54)/(5.2) = 1281.56 N

in init vector notation

F1 = 1.281*10^3 N (j)

B. Balancing forces in the vertical direction, we have F1 + F2 - Mg - mg = 0 ,

F2 = 270*9.81 + 42*9.81 - 1281.56 = 1779.16

F2 = 1.779*10^3 N (j)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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