A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 78.0 k

Question

A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 78.0 kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot located a distance ? = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 54.7 kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip.

(a) What is the appropriate analysis model for the beam before it begins to tip?

(b) Sketch a force diagram for the beam, labeling the gravitational and normal forces acting on the beam and placing the woman a distance x to the right of the first pivot, which is the origin.

This answer has not been graded yet.

Explanation / Answer

c)n1 is at maximum when x = 0 or the woman is directly over the left pivot

d) at the point of about to tip, n1 goes to zero

e) 0 = n2 – 54.7g - 78g

n2 = 132.7g

n2 = 132.7(9.8)

n2 = 1300.46 N

f) at the point of about to tip, n1 goes to zero so sum moments about the right hand pivot so that the only forces creating moments are the weight of the beam itself and the weight of the woman. I'll use mass instead of Force as all mass is affected by the constant gravity g. Assume CCW is positive

0 = 78(4 - 6/2) – 54.7(x - 4)

78(1) = 54.7x – 218.8

x = 5.426 m right of the left pivot

g) 1300.46(4) - 78(9.8)(3) – 54.7(9.8)(5.426)

? 0

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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