Three particles, each with a mass of 0.25 kg, are located at ( - 4.0 m, 0), (2.0 m, 0), and (0, 3.0 m) and are acted on by forces, F1=(-3.0N)y, F2=(5.0N)y, and F3=(4.0N)x respectively. Find the acceleration (magnitude and direction) of the center of mass of the system. [Hint: Consider the components of the acceleration.]
2006-07-22 15:18:05 · 4 answers · asked by jamey.moore@sbcglobal.net 1 in Science & Mathematics ➔ Physics
1.0 First, find the center of mass for the entire system.
1.1 The formula is
1.1.2 for the x-coordinate:
xcm = 1/M n m xi
1.1.3 for the y-coordinate:
yCM = 1/M n m yi
1.2.0 Calculating
1.2.1 for M: .25 + .25 + .25 /3 = .25
1.2.1 for x:
xcm = 4 (.25 * -4 + .25 * 2 + .25 *0) = - 2.0
1.2.2 for y:
ycm = 4 (.25 * 3) = 3
1.3.0 Center of Mass is: (-2,3)
2.0 A Newton is a measure of force.
2.0.1 Since we have the units of force and the mass of the bodies, we can find the accelerations.
2.1 1 Newton is 1 kilogram ( 1 meter/second/second)
2.1.1 This is the relation F = ma in metric units.
2.2 Calculating the acceleration for B1:
-3 = .25 (a) = - 12 meters/sec2
2.3 Calculating the acceleration for B2:
5 = .25(a) = 20 meters/ sec2
2.4 Calculating the acceleration for B3:
4 = .25(a) = 16 meters/ sec2
2.5 Adding the accelerations for B1, B2 and B3, we discover that the Center of Mass is accelerating by 16 meters/ sec2 in the x-direction, and 8 meters/ sec2 in the y-direction.
2006-07-22 18:03:59 · answer #1 · answered by Anonymous · 0⤊ 0⤋
To find the acceleration of the CM, you can treat the forces as if they work on the system as a whole.
The net force on the system (in vectors) is
[1] ... Fnet = (0, -3) + (0, 5) + (4, 0) = (4, 2) Newton
The mass of the system is m = 3 x 0.25 = 0.75 kg.
The acceleration is
[2] ... a = Fnet / m = (4, 2) / 0.75 = (5.33, 2.67) m/s2.
The direction is given by
[3] ... th = inv tan (2.67 / 5.33) = inv tan 0.5 = 26.6 degrees
and the magnitude by
[4] ... |a|^2 = 5.33^2 + 2.67^2 = 35.56, |a| = 5.96 m/s2.
2006-07-23 03:50:42 · answer #2 · answered by dutch_prof 4 · 0⤊ 0⤋
The center of mass position is the weighted average of the positions. The acceleration of the center of mass is the weighted average of the accelerations.
2006-07-23 02:15:04 · answer #3 · answered by Benjamin N 4 · 0⤊ 0⤋
I have the direction as 30º (relative to x-axis) and 4.472 N (more precisely 2 * sqrt(5) as the force
Hopefully, the acceleratation A = F / (m1 + m2 + m3) =
apprx 5.963 m/ sec^2
Haven't done one of these in a LONG time ... hope it's OK
2006-07-23 00:28:05 · answer #4 · answered by Julia C 4 · 0⤊ 0⤋