How do I find acceleration of two block.

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Homework Help Overview

The discussion revolves around finding the acceleration of two blocks connected by a string, with one block on an incline. The problem involves applying Newton's second law to analyze the forces acting on each block, including tension, friction, and gravitational components.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the forces acting on each block and how to apply Newton's second law to derive equations for tension and acceleration. There are attempts to clarify the roles of friction and gravity in the equations.

Discussion Status

Participants are actively engaging with the problem, sharing their attempts and clarifying concepts. Some have provided guidance on how to set up the equations, while others are exploring the implications of their calculations and the relationships between the forces involved.

Contextual Notes

There are references to specific values obtained from previous parts of the problem, and participants are questioning the validity of using those values in the current context. The discussion also highlights the need to consider the effects of acceleration on tension in the system.

cheater1
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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.
 
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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.
 
Doc Al said:
Combine those two equations and you can solve for both tension and acceleration.

This is the step I am stuck on. What Fb= 22.007N, Fc=30.08N, What equation are you referring to?
 
cheater1 said:
This is the step I am 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.
 
Doc Al said:
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 ?
 
cheater1 said:
Fc-Fb=F and F=ma ?
One step at a time. What forces act on block B?
 
Doc Al said:
One step at a time. What forces act on block B?

Friction and the Tension of the string.
 
cheater1 said:
Friction and the Tension of the string.
That's two of them. Don't forget gravity. Now apply Newton's 2nd law: ∑F = ma.
 
Doc Al said:
That's two of them. Don't forget gravity. Now apply Newton's 2nd law: ∑F = ma.

Tension - friction= ma?
 
  • #10
cheater1 said:
Tension - friction= ma?
Don't forget gravity.
 
  • #11
Doc Al said:
Don't forget gravity.
huh? gravity is a force?

so tension - friction - sin(36.9)(9.8) =ma ?
 
  • #12
cheater1 said:
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
Doc Al said:
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
 
Last edited:
  • #14
cheater1 said:
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
Doc Al said:
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
cheater1 said:
∑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
Doc Al said:
Good!

No. The tension will be different once things accelerate.

Then how is it possible to solve for tension?
 
  • #18
cheater1 said:
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
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
cheater1 said:
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
Doc Al said:
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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