the problem reads this "A person with mass m1 = 62 kg stands at the left end of a uniform beam with mass m2 = 93 kg and a length L = 3.5 m. Another person with mass m3 = 65 kg stands on the far right end of the beam and holds a medicine ball with mass m4 = 10 kg (assume that the medicine ball is at the far right end of the beam as well). Let the origin of our coordinate system be the left end of the original position of the beam as shown in the drawing. Assume there is no friction between the beam and floor.
What is the new x-position of the person at the left end of the beam? (How far did the beam move when the ball was throw from person to person?) "
I know we have to do something like 1.822826=[(60+100d]+93(1.75+d)+65(3.5+d)/230
(btw 1.822826 is the location of center of mass)
I keep getting d=.1271928947 but it's wrong, can someone double check my work please
the error could be related to significant figures
... just because the calculator shows them, doesn't mean they are there
all of the data in the problem is 2 sig fig
... except for the mass of the ball, which is one sig fig
assuming the ball is exactly 10 kg
... the best the answer can be is two sig fig
... yours looks like 10 ... like a calculator display
I also tried cutting it down to .13 but that was wrong, the website that I use sometimes wants us to be 99%accurate so it'll want more numbers than needed
I think this might be a calculation error on my part
total mass = 231
original cg = [93*1.75 + 75*3.5 ]/231 = 1.841
new cg = [93*1.75 + 65*3.5]/231 =1.689
so the rod moved 0.152
thank you damon, I was using the wrong numbers :)
You are welcome.
Sure, let's double-check your work!
To solve this problem, we can use the principle of conservation of linear momentum. Since there is no external force acting on the system, the total linear momentum before and after the ball is thrown must be the same.
Let's consider the left end of the beam as the reference point for our coordinate system. Initially, the person at the left end of the beam has a mass of 62 kg and is located at x = 0. The person on the right end has a mass of 65 kg and is located at x = 3.5 m.
When the ball is thrown from person to person, the system's total linear momentum is conserved. We can write the equation as follows:
(m1)(0) + (m2)(0) + (m3)(3.5) + (m4)(3.5 + d) = (m1 + m2 + m3 + m4)(1.822826)
Substituting in the given values, we have:
(62)(0) + (93)(0) + (65)(3.5) + (10)(3.5 + d) = (62 + 93 + 65 + 10)(1.822826)
Simplifying the equation:
(65)(3.5) + (10)(3.5 + d) = (230)(1.822826)
227.5 + 35 + 10d = 419.59538
Now, we can solve for d:
10d = 419.59538 - 227.5 - 35
10d = 156.09538
d = 15.609538
So, the correct value for d should be approximately 15.609538 meters.
To find the new x-position of the person at the left end of the beam, we need to calculate the displacement of the beam when the ball was thrown from person to person.
Let's analyze the problem step by step by considering the conservation of momentum and the conservation of angular momentum:
1. Conservation of momentum:
Since there is no external force acting on the system, the total momentum before and after the ball is thrown must be the same.
Before the ball is thrown, the total initial momentum is zero because the system is at rest (assuming no initial velocity). Therefore, we can write the equation:
m1v1 + m2v2 + m3v3 + m4v4 = 0
(Here, v1, v2, v3, and v4 are the velocities of persons 1, 2, 3, and the medicine ball respectively.)
2. Conservation of angular momentum:
Since there is no torque acting on the system, the total angular momentum before and after the ball is thrown must be the same.
Before the ball is thrown, the total initial angular momentum is also zero, considering the system is at rest. Therefore, we can write the equation:
m1x1v1 + m2x2v2 + m3x3v3 + m4x4v4 = 0
(Here, x1, x2, x3, and x4 are the x-positions of persons 1, 2, 3, and the medicine ball respectively.)
Now, we can substitute the given values into the equations:
m1 = 62 kg
m2 = 93 kg
m3 = 65 kg
m4 = 10 kg
L = 3.5 m (length of the beam)
Next, we need to express the velocities of each person and the medicine ball in terms of the displacement, d:
v1 = 0 m/s since person 1 is standing at rest
v2 = (d / t) m/s (velocity of person 2, where t is the time it takes for the ball to travel from person 2 to person 3)
v3 = (d / t) m/s (velocity of person 3 when receiving the ball)
v4 = 0 m/s since the medicine ball is initially at rest before being thrown.
By substituting these values into the momentum equation (from step 1) and the angular momentum equation (from step 2), we can solve for the displacement, d.
It seems that you've attempted the calculations, but you are receiving a different value for d. To find where the error might lie, double-check your calculations and make sure you have correctly accounted for all the variables and their interactions.
By rechecking your work and recalculating, you can determine whether you made an error in your calculations or if there might be a transcription mistake in the problem given.
If you provide the detailed steps of your calculation, we can help you pinpoint any mistakes and assist you in finding the correct value for d.