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I have a test coming up and I want to be able to understand the following problem:
Two crates are on a table crate 1 is attached by string A to crate 2 string B

A large crate (crate 1) has a mass of 4kg and the smaller crate (crate 2) has a mass of 2kg. The coefficient of kinetic friction between each crate and the wooden table is 0.1. A hanging (string C) weight of mass 1 kg causes both crates (crate 1 and crate 2) to accelerate to the right. What's the acceleration of the block and the tension T1 and T2 in strings A and B?

My attempt:
I think I would have to set up three different equations to find acceleration?


F=ma

Mass 1:

T-0.1(+9.8)=4a?
so T-.98=4a


Mass 2:
-T1+T2-.98=2a?

I'm really confused with trying to set up this equation.
I put -T1 because it's acting to the left and +T2 because it's
acting to the right.
-.98 because friction moves in the opposite direction of motion
so -1.0(9.8)=-.98


Mass 3:
1g - T2 = 1a?

I would appreciate any help, thanks!

I think I can help! The best thing to do for these is to draw a force diagram for each box to start with. Lets find out the normal force first; n1=m1(g), n2=m2(g). From that we find friction fk1=u(n1) fk2=u(n2).(u is the coefficient of friction). Ok, acceleration; Fnet=m3(g)-fknet, Fnet=ma, ma=m3(g)-fknet, a=(m3(g)-fknet)/m. m=m1+m2+m3. Now for the tensions and this part I'm not 100 percent on but I think it's right(unknown008 will correct me if I'm wrong). T2=Fnet-fknet, T1=T2-fk1. Do you have the correct answers in the back of the book or something? I'm curious myself!

OK, I tried solving for a but I don't think I got the correct answer, can you please check, thanks!

n1=4(9.8)=39.2
n2=2(9.8)=19.6
fk1=0.1(39.2)=3.92
fk2=0.1(19.6)=1.96
a=1(9.8)-5.88/39.2+19.6+9.8=0.0571

Is fknet 5.88 I got that number by adding 3.92 + 1.96
Also is m3(g)=1(9.8) or 1(-9.8) since acceleration is downward?
When I'm adding m1+m2+m3 is the following correct?
m1=39.2?
m2=19.6?
m3=9.8 or -9.8?

Sorry for all the questions but I need to know if I'm doing this correctly, sorry there are no answers
in the back of the book :/

Well... all those numbers in a single paragraph!

I'll do it.

Consider crate 3 (was that a typing error in your question?).

It's mass drives the whole system. Hence, we get:

mg - T_2 = ma

(1)(9.8) - T_2 = (1)(a)

9.8 - T_2 = a

Consider crate 2 only.

T_2 - T_1 - F_f = ma

T_2 - T_1 - (0.1)(2)(9.8) = (2)(a)

T_2 - T_1 - 1.96 = 2a

Consider crate 1 only;

T_1 - F_f = ma

T_1 - (0.1)(4)(9.8) = 4a

T_1 - 3.92 = 4a

For friction, remember that F_f = \mu R = \mu mg
------------------------
9.8 - T_2 = a

T_2 - T_1 - 1.96 = 2a

T_1 - 3.92 = 4a

Now can you solve them? :)

I get:

T_1 = 6.16 N
T_2 = 9.24 N
a = 0.56 m/s^2

Darn! I told you he would correct me but I can't be totally wrong. I have a football game to go to, we are playing Arkansas, but I'm coming back to this later.

Yeah I stand corrected. I started completely wrong because I wasn't taking into account the acceleration of the system which changes the tensions. Oh well, I just solved it correctly and it was exactly how you solved it. Thanks unknown008! Oh by the way we whooped Arkansas 65-43 haha! CAM NEWTON FOR HEISMAN!

Thank you!!