# Two blocks connected by string with an angled force applied

1. Oct 29, 2008

### blakester

A 3.5 kg block (m1) and a 1 kg block (m2) are connected by a string and are pushed across a horizontal surface by a force applied to the 1 kg block as shown below. The coefficient of kinetic friction between the blocks and the horizontal surface is 0.3. If the magnitude of vector F is 27 N, what is the tension in the string that connects the blocks? (θ = 20)

http://www.webassign.net/hrw8/6-p-096.gif

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Ok so my FBD say for block m1:
x: t-fk1= m1a
y: N1 - m1g = m1a(0)

which gives me N1 = 34.3

FBD for block m2:
x:Fcos[the]-t-fk2=m2a
y:n2-m2g-fsin[the]=m2a(0)

which gives me N2 = 13.904 ( = m2g+fsin20)

and fk1=10.29 and fk2 = 4.171 i know because I have the N's and the kinetic friction coefficients.

so with all that!! i did

T= fk1+m1a (from first eq for x of m1)

so i sub that into the m2's x equation

Fcos20-fk1-m1a-fk2=m2a
12cos20-10.29-3.5a-4.171 = 1a
a = -.7077

which is weird cause i didnt think there would be a negative acceleration.

Then i plugged the acceleration into the first eq for x on block 1
T-fk1=m1a which gave me 7.81305 for T but thats apparently the wrong answer, so if anyone can help find what i did wrong it would be greatly appreciated.

I have another problem like this i'm stuck on and i think if i figure out this one i should be able to get the next one too. Thanks for lookin

2. Oct 29, 2008

### PhanthomJay

Your method and equations looks OK after a cursory look, but I note that the value of F is given as 27 and you used 12. 12 isn't enough to make the system move! Correct the typo and you should be good.

3. Oct 29, 2008

### blakester

oh man haha, why would i do that! thanks for looking it over i got it right now!!!