Solve Kinematics Problem: Find Acceleration for 2.7kg Blocks on Inclined Pulley

In summary, the block on the incline has a weight of 2.7 kgs and has a coefficient of kinetic friction of .15. The block on the incline experiences an acceleration of 4.3 m/s^2 when the cable connecting the two blocks is pulled.
  • #1
pb23me
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0

Homework Statement


there are two blocks on a pully connected by a cable. one is on an incline and the other hangs in the air-see attatched photo. M1 = M2= 2.7kgs ; theta =34 ; coeffcient kinetic friction=.15 find accleration


Homework Equations


Fmax=N(coefficient kinetic friction)
Fnet=ma


The Attempt at a Solution

i assumed block B was going down, wrote down three Fnet equations.Fnet1y=mgcos(theta)-mgcos(theta)=0 ; Fnet1x=T-mgcos(theta)(coefficient kinetic fr)=Ma ; F2=mg-T=Ma therefore T=mg-ma plugged it in the other equation and solved for acceleration. i got 4.3m/s^2 I am curious about solving for the force of friction... i thought it was N(coefficient static fr) in this case i have N=mgcos(theta) is this correct?
 

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  • #2
hi pb23me! :smile:

(have a theta: θ and a mu: µ and try using the X2 and X2 icons just above the Reply box :wink:)

Can you write it out more clearly? It's difficult to check as it is.

(But I think you've left out the component of gravity along the slope.)

And yes, there's no acceleration perpendicular to the slope, so F = ma = 0 in that direction gives you N - mgcosθ = 0. :wink:
 
  • #3
well my only real question is what is the force due to friction? is it µ*mgcos[tex]\theta[/tex] or µ*mgsin[tex]\theta[/tex]
or is this equation correct ∑Fx=T-mg(cosθ)µ this is the equation for the block on the incline
 
Last edited:
  • #4
1)∑Fy=mgcosθ -mgcosθ=0
2)∑Fx=T-mg(cosθ)µ=ma
3)∑F=mg-T=ma
1&2 are for the block on the incline
 
  • #5
pb23me said:
well my only real question is what is the force due to friction? is it µ*mgcos[tex]\theta[/tex] or µ*mgsin[tex]\theta[/tex]

It's µN, whatever that happens to be. :wink:
or is this equation correct ∑Fx=T-mg(cosθ)µ this is the equation for the block on the incline

As I said before, you've left out the component of mg along the slope.

(not the friction force, the gravity itself, which would be there even if the slope was smooth)
 
  • #6
oh are you telling me that gravity is acting dwn and to the rght? i think i see what u mean
 
  • #7
It's what gravity does best! :wink:

I'm off to bed now …

goodnight! :zzz:​
 
  • #8
thanx for the help:)
 

What is kinematics?

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause the motion.

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a measure of how quickly an object's speed or direction changes.

How do I solve a kinematics problem?

To solve a kinematics problem, you will need to use equations that relate position, velocity, acceleration, and time. You will also need to consider the initial conditions of the objects in motion and any external forces acting on them.

What is the formula for finding acceleration in a kinematics problem?

The formula for finding acceleration in a kinematics problem is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

How do I find acceleration for 2.7kg blocks on an inclined pulley?

To find acceleration for 2.7kg blocks on an inclined pulley, you will need to consider the forces acting on the blocks, such as the weight of the blocks and the tension in the rope. You can then use the formula a = (Fnet / m), where a is acceleration, Fnet is the net force acting on the blocks, and m is the mass of the blocks. Solving for a will give you the acceleration of the blocks on the inclined pulley.

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