Frictionless Boxes: Finding Tension in Connecting Cord

[jarny]

Homework Statement

a box of mass m2 = 1.2 kg on a frictionless plane inclined at angle θ = 26°. It is connected by a cord of negligible mass to a box of mass m1 = 3.3 kg on a horizontal frictionless surface. The pulley is frictionless and massless. (a) If the magnitude of the horizontal force F-> is 3.7 N, what is the tension in the connecting cord? (b) What is the largest value the magnitude of F-> may have without the connecting cord becoming slack?

Homework Equations

Fnet=m*a->
T-m*g*sin(theta)=ma not sure if this applies here


link to picture: http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c05/fig05_56.gif

The Attempt at a Solution

Ok, I used F_g=m*a to find that the Force of gravity was 32.34N. To find the horizontal component I used F_g*tan(theta) to get 15.7733N going horizontally. This means that the tension in the rope is 15.7733N the opposite way towards m1. m1 has a f-> of3.7 so 15.7733-3.7=12.0733 and for b. it would be 15.7733 to be match the opposing force...I don't think this is right though, can someone help verify this or if it is wrong point me in the right direction? Thanks alot

 

[djeitnstine]

Draw a free body diagram T-mgsin \theta = ma has everything to do with it

 

[jarny]

Wait would the objects have different acceleration?

 

[jarny]

So would I have T-1.2*9.8*sin(26)=1.2*a and T-3.3*9.8*sin(0)=3.3*a so would I substitute and solve for T? and where does f-> come in? is it (T-3.3) for equation two? thanks

 

[Kruum]

So would I have T-1.2*9.8*sin(26)=1.2*a and T-3.3*9.8*sin(0)=3.3*a so would I substitute and solve for T? and where does f-> come in? is it (T-3.3) for equation two? thanks

You've got the first equation right. But the second one needs correction. If you draw the free body diagram, you can see there are two forces that cancel each others out. And also you need to separate vertical and horizontal forces. And for the horizontal force "F->" make sure you got the directions right.

 

[jarny]

I have the free body diagram, the f-> is confusing me.

<br /> F_(g,m_1)=9.8*3.3=32.34 Newtons__________ <br />

T-mgsin \theta = ma

the horizontal force is f-&gt; and it is moving in the positive direction.

All that I see for x is :

Sum(F_x)=fcos \theta + -F_gsin \theta

So Sum(f_x)= 3.7*cos(0)-32.34*sin(0)=3.7N

So that leaves 3.7=3.3*a

3.7/3.3=1.12 m/s^2

Is that how to find acceleration or did I mess up again?
***Don't know why tex puts acceleration calculation twice

 

[Kruum]

I have the free body diagram, the f-> is confusing me.

<br /> F_(g,m_1)=9.8*3.3=32.34 Newtons__________ <br />

T-mgsin \theta = ma

You've got the right equation here, but you don't need the gravity for m₁ since there is an equal force in the opposite direction.

the horizontal force is f-&gt; and it is moving in the positive direction.

Yes.

All that I see for x is :

Sum(F_x)=fcos \theta + -F_gsin \theta

I'm assuming "-F_gsin \theta" is the horizontal force of m₁'s gravity. If you take a closer look, you should see that the horizontal force "f->" and "-F_gsin \theta" aren't actually going in the same direction. So you can't sum them.

So Sum(f_x)= 3.7*cos(0)-32.34*sin(0)=3.7N

So that leaves 3.7=3.3*a

3.7/3.3=1.12 m/s^2

Is that how to find acceleration or did I mess up again?

You have forgotten the tension of the rope. And your answer wouldn't have been right anyway.

I would suggest you start from scratch and slice the original image into half where the pulley is. Draw a free body diagram for both pictures and make sure you have all the necessary forces (gravity, normal force, tension and f->). Since you have two unknowns, you need two equations. From the other you can solve T and the other for a. And you don't have to have numerical answer, you can leave symbols as well. By the way, one of the equations is within this page.