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. (Submit a file with a maximum size of 1 MB.) This answer has not been graded yet. (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? YesNo
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. (Submit a file with a maximum size of 1 MB.) This answer has not been graded yet. (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? YesNo
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Question
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. (Submit a file with a maximum size of 1 MB.)
(b) Where is the woman when the normal force
x = m
(c) What is
N
(d) Use the force equation of equilibrium to find the value of
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?
This answer has not been graded yet.
(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?
YesNo
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