A beam resting on two pivots has a length of
L = 6.00 m
and mass
M = 87.0 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 = 54.0 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.
(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
part 2
A beam of length L and mass M rests on two pivots. The first pivot is at the left end, taken as the origin, and the second pivot is at a distance ℓ from the left end. A woman of mass m starts at the left end and walks toward the right end as in the figure below.
(a) When the beam is on the verge of tipping, find a symbolic expression for the normal force exerted by the second pivot in terms of M, m, and g. (Enter the magnitude.)
n2 =
(b) When the beam is on the verge of tipping, find a symbolic expression for the woman's position in terms of M, m, L, and ℓ.
x =
(c) Find the minimum value of ℓ that will allow the woman to reach the end of the beam without it tipping. (Use the following as necessary: m, M, and L.)
ℓmin =
i know its a loaded question but I'm super stuck. please help me
out!
I'd be happy to help you with these questions. Let's break them down one by one.
(b) Where is the woman when the normal force n1 is the greatest?
To find the position of the woman when n1 is the greatest, we need to consider the forces at play. When the beam is in equilibrium, the sum of the torques acting on it must be zero. The torque due to the woman's weight, Mg, can be calculated as Mg * x, where x is the woman's position from the left end of the beam. To maximize n1, we want to maximize the torque due to the woman's weight. This occurs when x is at its maximum value, which is the right end of the beam. Therefore, the woman is at position x = L.
(c) What is n1 when the beam is about to tip?
When the beam is about to tip, the sum of the torques acting on it is no longer zero. To calculate n1 in this case, we can use the following equation:
(n1 * l) + (M * g * L) = (m * g * L)
where l is the distance between the first pivot and the second pivot. Solving for n1, we get:
n1 = (m * g * L - M * g * L) / l
(d) Use the force equation of equilibrium to find the value of n2 when the beam is about to tip.
When the beam is about to tip, the sum of the forces acting on it is no longer zero. To calculate n2 in this case, we can use the following equation:
n1 + n2 = M * g
Substituting the value of n1 from the previous question, we get:
[(m * g * L - M * g * L) / l] + n2 = M * g
Solving for n2, we get:
n2 = M * g - (m * g * L - M * g * L) / l
(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.
To find the woman's position when the beam is about to tip, we can set up the torque equation around the second pivot point. The equation is:
(n1 * l) + (n2 * 0) = (m * g * x)
Substituting the values of n1 and n2 from parts (c) and (d), we get:
[(m * g * L - M * g * L) / l] * l + n2 * 0 = m * g * x
Simplifying:
(m * g * L - M * g * L) + 0 = m * g * x
Dividing both sides by m * g:
(L - M * L / (m * g)) = x
Substituting x = L:
(L - M * L / (m * g)) = L
Simplifying:
- M * L / (m * g) = 0
This equation tells us that the woman's position when the beam is about to tip is at x = L.
(f) Check the answer to part (e) by computing torques around the first pivot point.
To check the answer in part (e), we can set up the torque equation around the first pivot point. The equation is:
(n1 * 0) + (n2 * (L - l)) = (m * g * x)
Substituting the values of n1 and n2 from parts (c) and (d), we get:
0 + [M * g - (m * g * L - M * g * L) / l] * (L - l) = m * g * x
Simplifying:
(M * g) * (L - l) - (m * g * L - M * g * L) * (L - l) / l = m * g * x
Expanding and simplifying:
M * g * L - M * g * l - m * g * L + M * g * L - M * g * L + m * g * L = m * g * x
Simplifying further:
M * g * L - M * g * l - m * g * L + m * g * L = m * g * x
M * g * L - M * g * l = m * g * x
Dividing both sides by m * g:
(L - l) = x
Substituting x = L:
(L - l) = L
Simplifying:
- l = 0
This equation tells us that the woman's position when the beam is about to tip is at x = L.
Part 2:
(a) When the beam is on the verge of tipping, the normal force exerted by the second pivot can be calculated using the equation for torque equilibrium around the second pivot point:
(n1 * l) + (n2 * 0) = (m * g * x)
Since the beam is on the verge of tipping, the sum of torques must be zero, and the normal force exerted by the second pivot would be zero. Therefore, n2 = 0.
(b) When the beam is on the verge of tipping, the woman's position can be found using the equation for torque equilibrium around the second pivot point:
(n1 * l) + (n2 * 0) = (m * g * x)
Since n2 = 0 when the beam is on the verge of tipping, the equation simplifies to:
(n1 * l) = (m * g * x)
Solving for x, we get:
x = (n1 * l) / (m * g)
(c) To find the minimum value of l that will allow the woman to reach the end of the beam without it tipping, we need to consider the forces and torques at play. The sum of torques acting on the beam around the second pivot must be zero to prevent tipping. The equation for torque equilibrium around the second pivot point is:
(n1 * l) + (n2 * 0) = (m * g * x)
Since n2 is zero when the beam is on the verge of tipping, the equation simplifies to:
n1 * l = m * g * x
Substituting x = L (reaching the end of the beam), we have:
n1 * l = m * g * L
Solving for l, we get:
l = (m * g * L) / n1
Therefore, to find the minimum value of l, we need to find the maximum value of n1. You can refer back to part (c) to calculate n1 using the given values and substitute it into the equation above to find the minimum value of l.