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Find tension and acceleration of 2 blocks stacked

  • #1

Homework Statement



A m1=7.47 kg block is placed on top of a m2=11.5 kg block. A horizontal force of 80.3 N is applied to the 11.5 kg block causing it to slide. The 7.47 kg block is tied to the wall by a string. The co-efficient of kinetic friction between all surfaces is 0.172.

a) Determine the magnitude of the tension in the string.
b) Find the acceleration of the 11.5 kg.

Use this image as reference --> http://images.4chan.org/sci/src/1295658037261.png [Broken]

Homework Equations



a) T - Ff = 0

b) Fnet = ma


The Attempt at a Solution



a) Ff = T
(0.172)(9.8)(7.47) = 12.6 N

b) ?

I have tried part b) for a very long time, and something is going wrong, can someone help me please!
 
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Answers and Replies

  • #2
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The friction will be acting in forward direction on 7.47Kg block and backward on 11.5 kg block(opposite to 80.3N force)

So draw the FBD, draw the 2 forces and find tension
 
  • #3
tiny-tim
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Welcome to PF!

Hi ILoveTurtles!

Song of Solomon 2:12 The flowers appear on the earth; the time of singing is come, and the voice of the turtle is heard in our land!​

Welcome to PF! :wink:
A m1=7.47 kg block is placed on top of a m2=11.5 kg block. A horizontal force of 80.3 N is applied to the 11.5 kg block causing it to slide. The 7.47 kg block is tied to the wall by a string. The co-efficient of kinetic friction between all surfaces is 0.172.

b) Find the acceleration of the 11.5 kg.
Use good ol' Newton's second law (on the bottom mass only), Ftotal = ma …

show us what you have so far :smile:
 
  • #4
Okay, I thought it might be:

Fnet = ma
Fa - Ff = m2a
(80.3) - (12.6) = (11.5)a
a=5.89 m/s^2

But this is wrong! Am i missing a negative? or a whole force? so frustrating! lol!
 
  • #5
SammyS
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Okay, I thought it might be:

Fnet = ma
Fa - Ff = m2a
(80.3) - (12.6) = (11.5)a
a=5.89 m/s^2

But this is wrong! Am i missing a negative? or a whole force? so frustrating! lol!
Hello.

The 11.5 kg block sits on a surface. As stated in the original post:
The co-efficient of kinetic friction between all surfaces is 0.172.
The frictional force on the bottom block is significantly greater than 12.6 N. You only considered the frictional force the upper block exerts on the bottom block.

Find the Normal Force that the supporting surface exerts on the bottom block. Then find the frictional force the supporting surface exerts on the bottom block. Add this to the frictional force you did include.
 
  • #6
Hello.

The 11.5 kg block sits on a surface. As stated in the original post:

The frictional force on the bottom block is significantly greater than 12.6 N. You only considered the frictional force the upper block exerts on the bottom block.

Find the Normal Force that the supporting surface exerts on the bottom block. Then find the frictional force the supporting surface exerts on the bottom block. Add this to the frictional force you did include.
Oh! So I calculated the frictional force on the bottom block (19.38 N), but Im still confused how to incorporate it into the equation. Im getting thrown off by the fact that the tension is only on the top block and i dont know which forces to add and subtract.
Could it be:

80.3 - 19.4 - 12.6 = 11.5 a ?
a = 4.20 m/s^2
 
  • #7
tiny-tim
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Hi ILoveTurtles! :wink:

That's the correct basic principle, but you've calculated the friction force at the ground wrongly.

The friction force at the ground is the normal force times 0.172, and you've used the wrong normal force.

In questions like this, you need to draw a free body diagram for each block.

For example, the fbd fot the upper block would show four arrows, one for the tension, one for the friction, one for the weight (of the upper block), and one for the normal force (between the two blocks).

Tell us what you think the fbd for the lower block should look like. :smile:
 
  • #8
SammyS
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...
In questions like this, you need to draw a free body diagram for each block.
...
Hi ILoveTurtles!

I fully agree with tiny-tim. I, too, should have suggested a set of free-body diagrams. They help resolve a lot of issues!
 
  • #9
so the FBD for the lower block, would be horizontal arrow forward, friction backward, weight of both masses downward, normal upward AND tension backward? im so unsure about the tension....

so this is what i gather,

Fn = (m1 + m2)(9.8) = 186 N??
Ff = 186 x 0.172 = 32 N on lower block

Fnet = ma
Fh - Ff - T= m2a
Fh - (mu)(Fn) - T= m2a
80.3 - (0.172)(11.5 + 7.47)(9.8) - 12.6 = 11.5 a
a = 3.12 m/s^2

- im still not sure about subtracting the tension, is that the right value?
 
  • #10
tiny-tim
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so the FBD for the lower block, would be horizontal arrow forward, friction backward, weight of both masses downward, normal upward AND tension backward? im so unsure about the tension....
ohh, this is fairly bad :redface:

you do need to fully understand fbds, so let's take this slowly …

first, the fbd for a body (in this case, the lower block) only has forces on that body

second you seem to have missed out at least one force

this time, say where each arrow is (top, bottom, left right) :wink:
 
  • #11
:( hmm, okk

horizontal force = right
force of gravity = bottom
force of friction = left
force normal = top
tension = left

im missing a force :O i cant figure out which one that iss :(
 
  • #12
actually, there wouldnt be tension then, because that is not on the bottom block!
 
  • #13
I guess I would be missing the accelerating force? F=ma
 
  • #14
tiny-tim
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Hi ILoveTurtles! :wink:
:( hmm, okk

horizontal force = right
force of gravity = bottom
force of friction = left
force normal = top
tension = left

im missing a force :O i cant figure out which one that iss :(
how about force normal = bottom ? :smile:

also, you're missing one of the friction forces

and the friction force you have included, you've said which way it's pointing (left), but not where it's attached

every arrow not only has a direction, but usually (though not always) the point of application of the arrow (ie of the force) is also important
actually, there wouldnt be tension then, because that is not on the bottom block!
correct! :biggrin:
I guess I would be missing the accelerating force? F=ma
no such thing in a free body diagram … only put individual forces, not "total" forces

ok, draw the fbd with two normal forces (one top one bottom), two friction forces (one top one bottom), the weight of the lower block only, and the pulling force F …

give the different forces letters (so you can for example distinguish between the two normal forces), and then write out equations for each of the forces :smile:
 
  • #15
im confused why the friction is top and bottom....and am i making equations of the x and y components? because if that is the case, then there is only 1 force in the x direction (F)

i dont get it :(
 
  • #16
tiny-tim
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im confused why the friction is top and bottom
There's friction between the two blocks (so that's at the top of the lower block),

and friction between the lower block and the ground.
....and am i making equations of the x and y components? because if that is the case, then there is only 1 force in the x direction (F)
Yes, you'll be doing the x and y components separately.

But no, there's 3 forces in the x direction … F and the two friction forces. :wink:
 
  • #17
okk, i see what u mean, theres friction at the top and bottom surfaces, but which direction does each friction force move in? wouldnt they BOTH move opposite to the horizontal force? so would the x direction equation be:

F - Ff1 - Ff2 = m2a

or am i completely missing something.....because that cantt be right ughh
 
  • #18
tiny-tim
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F - Ff1 - Ff2 = m2a
Looks right to me :smile:

what's worrying you about it? :confused:

(next step of course is to find what Ff1 and Ff2 are)
 
  • #19
hey! okay! soo nowwww, i would thinkk,

Ff1 = (7.47)(9.8)(0.172) = 12.6 N
Ff2 = (11.5 + 7.47)(9.8)(0.172) = 32 N
80.3 - 12.6 - 32 = 11.5 a
a = 3.12 m/s^2 :( this is what i had earlier :(
 
  • #20
its right!!!!! the answer is 3.10 m/s^2 !! thank you soooooooooooo much for all your help tiny tim! i actually understanddd the question now lol!
 
  • #21
tiny-tim
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Ff1 = (7.47)(9.8)(0.172) = 12.6 N
Ff2 = (11.5 + 7.47)(9.8)(0.172) = 32 N
80.3 - 12.6 - 32 = 11.5 a
a = 3.12 m/s^2 :( this is what i had earlier :(
let's see …
so the FBD for the lower block, would be horizontal arrow forward, friction backward, weight of both masses downward, normal upward AND tension backward? im so unsure about the tension....

so this is what i gather,

Fn = (m1 + m2)(9.8) = 186 N??
Ff = 186 x 0.172 = 32 N on lower block

Fnet = ma
Fh - Ff - T= m2a
Fh - (mu)(Fn) - T= m2a
80.3 - (0.172)(11.5 + 7.47)(9.8) - 12.6 = 11.5 a
a = 3.12 m/s^2
Yes, the only difference is that you originally included the 12.6 as the tension (which isn't acting on the lower block) …

in this case it happens to be the same as the (upper) friction, but if for example it was a spring instead of a rope, then the upper block could move, and there would be a difference.

(It's a short-cut that happens to work in this case, but I'm pretty sure you'd lose marks in the exam if you used it. :wink:)

EDIT: oh, I missed your last post! :biggrin:

I hope you're familiar with fbds now! :wink:
 

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