How do I find acceleration of two block.

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Homework Statement



http://yfrog.com/0k43313183j

Homework Equations




The Attempt at a Solution



Ok i got a, b, and c. But i need help with d.

My attempt was to find the force of block b - Tension of block C

block B = sin(36.9)25 - the force on block b
cos(36.9)25*0.35 - which is the friction.

block C = 30.8N (which I got from part c)


Then I took the force on C-B which I got 22.787N but then tried to divide by the mass to get acceleration. This is there I'm confused. Please help me out and thanks in advance.
 

Answers and Replies

  • #2
Doc Al
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What forces act on block B? Apply Newton's 2nd law.
What forces act on block C? Apply Newton's 2nd law.

Combine those two equations and you can solve for both tension and acceleration.
 
  • #3
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Combine those two equations and you can solve for both tension and acceleration.
This is the step im stuck on. What Fb= 22.007N, Fc=30.08N, What equation are you referring to?
 
  • #4
Doc Al
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This is the step im stuck on. What Fb= 22.007N, Fc=30.08N,
What do those forces represent?

What equation are you referring to?
Each time you apply Newton's 2nd law, you'll get an equation. So you'll get one equation for block B and another for block C.
 
  • #5
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What do those forces represent?


Each time you apply Newton's 2nd law, you'll get an equation. So you'll get one equation for block B and another for block C.
Fc-Fb=F and F=ma ?
 
  • #6
Doc Al
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Fc-Fb=F and F=ma ?
One step at a time. What forces act on block B?
 
  • #7
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One step at a time. What forces act on block B?
Friction and the Tension of the string.
 
  • #8
Doc Al
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Friction and the Tension of the string.
That's two of them. Don't forget gravity. Now apply Newton's 2nd law: ∑F = ma.
 
  • #9
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That's two of them. Don't forget gravity. Now apply Newton's 2nd law: ∑F = ma.
Tension - friction= ma?
 
  • #10
Doc Al
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  • #11
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Don't forget gravity.
huh? gravity is a force?

so tension - friction - sin(36.9)(9.8) =ma ?
 
  • #12
Doc Al
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huh? gravity is a force?
Of course! The weight of the box.

so tension - friction - sin(36.9)(9.8) =ma ?
Almost. The weight of block B is given as 25 N. What are its components parallel and perpendicular to the incline? Express the friction force in terms of the normal force.
 
  • #13
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Of course! The weight of the box.


Almost. The weight of block B is given as 25 N. What are its components parallel and perpendicular to the incline? Express the friction force in terms of the normal force.
N= cos(36.9)(25)
Friction = (cos(36.9)25)(0.35)
Tension = 30.8N

∑F= t-f-n?

So i got 30.8-19.99-6.997=3.813N

and mass is 25/9.8=2.55kg

so in the end i get 3.813=2.55a

a = 1.495m/s^2
 
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  • #14
Doc Al
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N= cos(36.9)(25)
Friction = (cos(36.9)25)(0.35)
Good.
Tension = 30.8N
You won't know the tension until you solve for it. Where did you get this value from?

∑F= t-f-n?
Almost. Not "-n" but minus the parallel component of the weight.
 
  • #15
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Good.

You won't know the tension until you solve for it. Where did you get this value from?


Almost. Not "-n" but minus the parallel component of the weight.
∑F= t-f-parallel?

N= cos(36.9)(25)
Friction = (cos(36.9)25)(0.35)
parallel= sin(36.9)25
T= can't i use the value i got from part c) of the problem?
 
  • #16
Doc Al
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∑F= t-f-parallel?

N= cos(36.9)(25)
Friction = (cos(36.9)25)(0.35)
parallel= sin(36.9)25
Good!
T= can't i use the value i got from part c) of the problem?
No. The tension will be different once things accelerate.
 
  • #17
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Good!

No. The tension will be different once things accelerate.
Then how is it possible to solve for tension?
 
  • #18
Doc Al
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Then how is it possible to solve for tension?
First finish with block B: write out the equation for Newton's 2nd law.

Then do a similar analysis for block C. What forces act on it?

You'll end up with two equations and two unknowns.
 
  • #19
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ma=t-f-parallel (block b)

For block C there is only Tension and Weight. F=t-w, ma=t-w (block c), right?
 
  • #20
Doc Al
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ma=t-f-parallel (block b)
Good. What's the mass of block B?

For block C there is only Tension and Weight. F=t-w, ma=t-w (block c), right?
Good. Two points:
(1) Since the acceleration will be downward, take down as positive: W-T = ma.
(2) What's the mass of block C? (You already know its weight from part c.)
 
  • #21
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Good. What's the mass of block B?


Good. Two points:
(1) Since the acceleration will be downward, take down as positive: W-T = ma.
(2) What's the mass of block C? (You already know its weight from part c.)
Omg you did it!

Mass of B will be 25/9.8=2.55
Mass of C will be (the answer I got from part c), 30.8/9.8=3.14
so with the questions you gave me

w-t=ma and t-f-p=ma

i can t=w-ma and replace t in the 2nd equation

w-ma-f-p=ma, and replacing with what I already know.

30.8-3.14a-cos(36.9)(25)(0.35)-sin(36.9)25=2.55a

8.79=5.69a

a=1.54m/s^2

Thank you so much for you help, never could have done this without you!
 
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