View Full Version : Newton's Laws -- friction, tension, acceleration
buckybadger
Oct2-09, 12:49 AM
A block of mass m1 = 2 kg and a block of mass m2 = 3 kg are tied together and are pulled from rest across the floor by a force of Fp = 24 N. The coefficient of friction of the blocks with the floor is μ = 0.27.
a) What is the acceleration of the two blocks?
b) What is the tension in the string between the blocks?
I think that this can be solved by setting up two equations:
m1a = T - μm1g
and
m2a = Fp - T - μm2g
The unknowns in those two equations are exactly what I need -- acceleration and tension.
Anybody know how to solve this?!
What have you tried so far to solve them? Hint: Try adding them together.
buckybadger
Oct2-09, 10:20 AM
Yes, I tried adding the two together by solving for T and plugging that equation into the other. Unfortunately, after crunching the numbers, I did not come to the correct answer.
Yes, I tried adding the two together by solving for T and plugging that equation into the other. Unfortunately, after crunching the numbers, I did not come to the correct answer.
Show what you did.
(Just add the two equations. What happens to the T terms?)
buckybadger
Oct2-09, 10:43 AM
The two equations are:
m1a = T - μm1g
m2a = Fp - T - μm2g
I solved for T as such:
T = m1a + μm1g
Substituting this T value into the second equation, I got:
m2a = Fp - (m1a + μm1g) - μm2g
So,
3a = 24 - (2a + (0.27)(2)(9.8)) - (0.27)(3)(9.8)
Solving for a, I got 5a = 21.35, or a = 4.27 m/s^2. Unfortunately, I do not believe that this is the correct answer.
(Thanks for helping me with this!!)
So,
3a = 24 - (2a + (0.27)(2)(9.8)) - (0.27)(3)(9.8)
Solving for a, I got 5a = 21.35, or a = 4.27 m/s^2. Unfortunately, I do not believe that this is the correct answer.
No, not correct. Your equation is fine, just check your arithmetic. (Careful with parentheses and signs.)
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