Newton's Laws - friction, tension, acceleration

[buckybadger]

Newton's Laws -- friction, tension, acceleration

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 - μm2gThe unknowns in those two equations are exactly what I need -- acceleration and tension. Anybody know how to solve this?!

 

[Doc Al]

What have you tried so far to solve them? Hint: Try adding them together.

 

[buckybadger]

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.

 

[Doc Al]

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]

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!)

 

[Doc Al]

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.)