(LIFE SAVER QUICK) (LIFE SAVER QUICK) A beam resting on two pivots has a length

Question

(LIFE SAVER QUICK)

(LIFE SAVER QUICK) A beam resting on two pivots has a length of L= 6.00 m and mass M=78.0kg. The pivot under the left end exerts a normal force n1 on the beam, and the second pivot placed a distance l= 4.00 m from the left end exerts a normal force n2. A woman of mass m=53.5kg 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. (c) What is n1 when the beam is about to tip? N (d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip. N (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. m (f) Check the answer to part (e) by computing torques around the first pivot point. m

Explanation / Answer

Part C)

When the beam is about to tip, it will pivout about the point at n2, and therefore no force will be exterted at n1. The beam will break contact at the location of n1 so the force exerted is zero N.

Part D)

When the beam is about to tip, as stated in part c, n1 will not factor into the equation. There are two forces actibg down, the weight of the woman and the weight of the beam. There is only one force acting up and that is the force n2

So n2 = (mg) + (mg) = (78)(9.8) + (53.5)(9.8) = 1288.7 N

Part E)

Using the pivot at point n2, there will be only two torques acting. The weight of the woman will tend to rotate one direction, and the weight of the beam will tend to rotate in the other direction. Therefore those torques are exactly balanced

Since = Fd, we have

Fd(beam) = Fd(woman)

The weight of the beam acts at it center, and that center point is 1 m to the left of n2

(78)(9.8)(1) = (53.5)(9.8)(d)

d = 1.46 m to the right of the n2

You can say that is 5.46 m from one end of the beam, or 0.54 m from the other end. Whichever your solution wants you to enter

Part F)

Using a pivot point at n1 we have three torques. The weight of the beam and woman each tend to rotate one direction, and the force at n2 will rotate the other way, so

Fd(beam) + Fd(woman) = Fd(n2)

(78)(9.8)(3) + (53.5)(9.8)(x) = (1288.7)(4)

x = 5.46 m from point n1 which is the exact value we found from part E

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