A beam resting on two pivots has a length of L = 6.00 m and mass M = 77.0 kg. The pivot under the left end exerts a normal force n, on the beam, and the second pivot placed a distance { = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 57.0 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 e from the first pivot at a shorter distance than the length M of mass m g across aeam which is resting of the beam. stanc Tom the end oe Deam The end o beam and the firstvot at a shorter pi directly dista (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.) Choose File No file chosen (b) Where is the woman when the normal force n, is the greatest? X = m (c) What is n, when the beam is about to tip? N (d) Use the force equation of equilibrium to find the value of n, when the beam is about to tip. (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 =
A beam resting on two pivots has a length of L = 6.00 m and mass M = 77.0 kg. The pivot under the left end exerts a normal force n, on the beam, and the second pivot placed a distance { = 4.00 m from the left end exerts a normal force n2. A woman of mass m = 57.0 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 e from the first pivot at a shorter distance than the length M of mass m g across aeam which is resting of the beam. stanc Tom the end oe Deam The end o beam and the firstvot at a shorter pi directly dista (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.) Choose File No file chosen (b) Where is the woman when the normal force n, is the greatest? X = m (c) What is n, when the beam is about to tip? N (d) Use the force equation of equilibrium to find the value of n, when the beam is about to tip. (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 =
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Transcribed Image Text:A beam resting on two pivots has a length of L = 6.00 m and mass M = 77.0 kg. The pivot under the left end exerts a
normal force n, on the beam, and the second pivot placed a distance { = 4.00 m from the left end exerts a normal force
n,. A woman of mass m = 57.0 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 e from the first pivot at a shorter distance than the length
M
woman of mass m g across a/eam which is resting of the beam.
andmess Mand the
man is a distance x from the left end oAne beam. The
pivot is directly under the left end ohe beam and the
cond pivot is a distance from the first ivot at a shorter
n the lenh 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.)
Choose File No file chosen
(b) Where is the woman when the normal force n, is the greatest?
X =
m
(c) What is n, when the beam is about to tip?
N
(d) Use the force equation of equilibrium to find the value of n, when the beam is about to tip.
(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 =
(f) Check the answer to part (e) by computing torques around the first pivot point.
X =
Except for possible slight differences due to rounding, is the answer the same?
Yes
O No
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