A beam resting on two pivots has a length of L = 6.00 m and mass M = 89.0 kg.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 = 52.5 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 woman of mass m walking across a beam which is resting on two pivots. The beam is of length L and mass M and the woman is a distance x from the left end of the beam. The first pivot is directly under the left end of the beam and the second pivot is a distance ℓ from the first pivot at a shorter distance than the length of the beam. (a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin. (b) Where is the woman when the normal force n1 is the greatest? x = m (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. x = m (f) Check the answer to part (e) by computing torques around the first pivot point. x = m Except for possible slight differences due to rounding, is the answer the same? Yes or No

Principles of Physics: A Calculus-Based Text

5th Edition

ISBN:9781133104261

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter10: Rotational Motion

Section: Chapter Questions

Problem 28P: A uniform beam resting on two pivots has a length L = 6.00 m and mass M = 90.0 kg. The pivot under...

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A beam resting on two pivots has a length of L = 6.00 m and mass M = 89.0 kg.The pivot under the left end exerts a normal force

n₁on 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 = 52.5 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 woman of mass m walking across a beam which is resting on two pivots. The beam is of length L and mass M and the woman is a distance x from the left end of the beam. The first pivot is directly under the left end of the beam and the second pivot is a distance ℓ from the first pivot at a shorter distance than the length of the beam.

(a) Sketch a free-body diagram, labeling the gravitational and normal forces acting on the beam and placing the woman x meters to the right of the first pivot, which is the origin.

(b) Where is the woman when the normal force

n₁

is the greatest?
x = m

(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.
x = m

(f) Check the answer to part (e) by computing torques around the first pivot point.
x = m

Except for possible slight differences due to rounding, is the answer the same?

Yes or No

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