One 2.00 kg paint bucket is hanging by a massless cord from another 2.00 kg paint bucket also hanging by a massless cord.
If the buckets are at rest what is the tension in the lower cord?
What is the tension in the upper cord?
If the two buckets are pulled upward with an acceleration of 1.48 m/s^2 by the upper cord, what is the tension in the lower cord?
Calculate the tension in the upper cord.
I have no idea how to go about solving these. What are the formulas to use?
2006-09-14 09:14:15 · 3 answers · asked by beautyqueenjustine 3 in Science & Mathematics ➔ Physics
Much simpler than they appear.
The formula in all cases is F=ma.
F is the force
m is the mass
a is acceleration, and gravity is about 9.8m/s^2.
1. The weight on the bottom cord is only the bottom bucket.
F = 2 x 9.8 = 19.6
2. the total weight on the top cord is the total weight of both buckets, ie 4kg.
F = 4 x 9.8 = 39.2
3. The upward acceleration is adding to gravity, so now "a" = 1.48 + 9.8 = 11.28.
Upper cord is still supporting both buckets so...
F = 4 x 11.28 = 45.12.
Those values for F are in Kg m / s^2, called "kilograms force". Often, force is represented in Newtons (N). the conversion is about 1 Kg m / s^2 = 9.8N. Just multiply the above results by 9.8 for an approximate value in Newtons.
2006-09-14 09:30:06 · answer #1 · answered by Dan C 2 · 0⤊ 1⤋
Start with F=ma (Force = Mass * acceleration)
g=9.8 m/s^2 (acceleration due to gravity)
1 N = 1 kg * 1 m/s^2.
You wanted formulas, not the answers.
2006-09-14 16:22:00 · answer #2 · answered by An electrical engineer 5 · 0⤊ 0⤋
For both scenarios start with the "free-body" diagram.
1st case, no acceleration:
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The free-body diagram of the upper bucket should show the tension of the upper cord (T1) pointing up, and the tension in the lower cord (T2) PLUS mg of the upper bucket pointing down.
The free-body diagram of the lower bucket should show T2 pointing up and mg of the lower bucket pointing down.
The 2 free-body diagrams lead to the equations,
T1 - T2 - mg = 0 and T2 - mg = 0
giving,
T1 = 2mg and T2 = mg
2nd case, acceleration:
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Same procedure as above EXCEPT now the sum of ALL forces does NOT equal zero. This will lead to the equations,
T1 - T2 - mg = ma and T2 - mg = ma
giving,
T1 = 2m(a + g) and T2 = m(a + g)
2006-09-14 16:53:06 · answer #3 · answered by entropy 3 · 0⤊ 0⤋