A beam resting on two pivots has a length of L = 9.00 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 4.00 m from the left end exerts a normal force n2. A woman of mass m= 48.0 kg steps onto the left end of the beam and begins walking to the right as in the figure. The goal is to find the woman’s position when the beam begins to tip. (d) Use the torque equilibrium equation, with torques computed around the second pivot point and find the woman’s position when the beam is about to tip
Why did the beam go to therapy?
Because it had a lot of "torque" in its life!
But in all seriousness, let's get to the question.
To find the woman's position when the beam is about to tip, we can use the torque equilibrium equation. The torque equation states that the sum of the torques acting on an object must be equal to zero for equilibrium.
In this case, we'll use the second pivot point as the fulcrum to calculate the torques. When the beam is about to tip, the torques on both sides of the pivot point will be equal.
The torque due to the woman can be calculated as follows:
Torque_woman = (m * g * x)
The torque due to the beam can be calculated as follows:
Torque_beam = (M * g * (L - x))
Setting the torques equal to each other:
(m * g * x) = (M * g * (L - x))
Now we can solve for x, the woman's position when the beam is about to tip:
(m * g * x) = (M * g * (L - x))
48.0 kg * 9.8 m/s^2 * x = 89.0 kg * 9.8 m/s^2 * (9.00 m - x)
Now, let me calculate that for you... drum roll, please!
Calculating... calculating...
The woman's position when the beam is about to tip is approximately x = 6.60 meters from the left end.
Hope that helps! If you need more assistance, just let me know.
To find the woman's position when the beam is about to tip, we can use the torque equilibrium equation around the second pivot point.
The torque equilibrium equation states that the sum of the torques acting on an object must be equal to zero.
First, let's calculate the torques acting on the beam and the woman.
1. Torque due to the beam's weight:
The weight of the beam acts at its center of mass, which is 4.5 m from either pivot point. The torque due to the beam's weight is given by:
T_beam = M * g * (L/2)
2. Torque due to the woman's weight:
Since the woman is standing on the left end of the beam, her weight acts at a distance L from the second pivot point. The torque due to the woman's weight is given by:
T_woman = m * g * L
Now, let's set up the torque equilibrium equation:
T_beam + T_woman = 0
Since the beam is about to tip, the sum of the torques must be zero.
Substituting the values we know, we have:
M * g * (L/2) + m * g * L = 0
Now we can solve for L, the position of the woman when the beam is about to tip.
Plugging in the given values: M = 89.0 kg, m = 48.0 kg, g = 9.8 m/s^2, and L = 9.0 m, we have:
89.0 kg * 9.8 m/s^2 * (9.0 m/2) + 48.0 kg * 9.8 m/s^2 * 9.0 m = 0
Simplifying the equation:
(89.0 kg * 9.8 m/s^2 * 4.5 m) + (48.0 kg * 9.8 m/s^2 * 9.0 m) = 0
(3913.8 N*m) + (4243.2 N*m) = 0
8157.0 N*m = 0
Since the equation simplifies to 0 = 0, it means that the beam is in equilibrium and will not tip.
Therefore, there is no specific position for the woman when the beam is about to tip.
To find the woman's position when the beam is about to tip, we need to analyze the torque equilibrium equation around the second pivot point.
Torque (or moment) is the rotational analogue of force and is given by the product of force and the distance from the pivot point. The torque equilibrium equation states that the sum of the torques acting on an object must be zero for it to be in rotational equilibrium.
Let's denote the distance of the woman from the second pivot point as x. At this point, the entire weight of the beam and the woman will act as a downward force at the center of mass, which is located at L/2 distance from the left end of the beam.
Now, let's calculate the torques acting on the beam.
1. Torque due to the weight of the beam:
The weight of the beam acts downward at its center of mass, which is at L/2. The torque due to this force is (M*g)*(L/2), where g is the acceleration due to gravity.
2. Torque due to the weight of the woman:
The weight of the woman acts downward at her location, which is x from the left end of the beam. The torque due to this force is m*g*x.
3. Torque due to the normal force n1:
The normal force n1 acts at the left pivot point, which is at 0 distance from the second pivot point. Since the distance is 0, the torque due to this force is zero.
4. Torque due to the normal force n2:
The normal force n2 acts at the second pivot point itself. Again, since the distance is 0, the torque due to this force is zero.
As per the torque equilibrium equation, the sum of these torques must be zero:
(M*g)*(L/2) + m*g*x = 0
Simplifying this equation:
(M*g*L)/2 = -m*g*x
Dividing both sides by g:
(M*L)/2 = -m*x
Solving for x, the woman's position:
x = -(M*L)/(2*m)
Substituting the given values:
M = 89.0 kg
L = 9.00 m
m = 48.0 kg
x = -(89.0 * 9.00)/(2 * 48.0)
x ≈ -12.86 m
The negative sign indicates that the woman's position is to the left of the second pivot point. Thus, the woman's position, when the beam is about to tip, is approximately 12.86 m to the left of the second pivot point.