Frictionless Boxes: Finding Tension in Connecting Cord

In summary, the conversation is discussing a problem involving a box of mass 1.2 kg on a frictionless inclined plane connected by a cord to a box of mass 3.3 kg on a horizontal frictionless surface. The pulley is also frictionless and has no mass. The conversation centers around two questions: (a) determining the tension in the connecting cord when a horizontal force of 3.7 N is applied, and (b) finding the maximum value of F-> without the cord becoming slack. The solution involves using the equations Fnet=m*a-> and T-m*g*sin(theta)=ma, as well as drawing a free body diagram and separating the vertical and horizontal forces. The conversation also discusses the confusion over the
  • #1
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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
 
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  • #2
Draw a free body diagram [tex]T-mgsin \theta = ma[/tex] has everything to do with it
 
  • #3
Wait would the objects have different acceleration?
 
  • #4
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
 
  • #5
jarny said:
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.
 
  • #6
I have the free body diagram, the f-> is confusing me.

[tex]
F_(g,m_1)=9.8*3.3=32.34 Newtons__________
[/tex]

[tex] T-mgsin \theta = ma[/tex]

the horizontal force is [tex] f-> [/tex] 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

[tex] 3.7/3.3=1.12 m/s^2 [/tex]

Is that how to find acceleration or did I mess up again?
***Don't know why tex puts acceleration calculation twice
 
  • #7
jarny said:
I have the free body diagram, the f-> is confusing me.

[tex]
F_(g,m_1)=9.8*3.3=32.34 Newtons__________
[/tex]

[tex] T-mgsin \theta = ma[/tex]

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 [tex] f-> [/tex] 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

[tex] 3.7/3.3=1.12 m/s^2 [/tex]

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.
 

1. What is the purpose of studying frictionless boxes and tension in connecting cord?

The purpose of studying frictionless boxes and tension in connecting cord is to better understand the principles of mechanics and how forces act on objects. This knowledge can then be applied to real-world situations, such as designing structures or predicting the movement of objects.

2. How is tension calculated in a system of frictionless boxes and connecting cord?

Tension can be calculated by using the equation T = ma, where T is the tension in the cord, m is the mass of the box, and a is the acceleration of the box. In a frictionless system, the tension in the cord will be equal to the weight of the box, since there is no other force acting on it.

3. What are some factors that can affect the tension in a system of frictionless boxes and connecting cord?

The tension in a system of frictionless boxes and connecting cord can be affected by the mass of the boxes, the angle of the cord, and the acceleration of the system. Other factors, such as air resistance and the elasticity of the cord, may also play a role.

4. Can the tension in a system of frictionless boxes and connecting cord ever be greater than the weight of the boxes?

No, in a frictionless system, the tension in the cord will always be equal to the weight of the boxes. This is because there are no other forces acting on the boxes to increase the tension.

5. How can the concept of tension in a frictionless system be applied to real-world scenarios?

The concept of tension in a frictionless system can be applied to real-world scenarios such as designing bridges, building cranes, and predicting the movement of objects on pulleys or cables. Understanding tension and how it affects the movement of objects is crucial in these situations to ensure structural stability and safety.

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