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

AI Thread Summary
The discussion centers on solving a kinematics problem involving two 2.7 kg blocks connected by a pulley, with one block on an incline at a 34-degree angle and the other hanging. The user attempts to find the acceleration and expresses confusion about calculating the force of friction, questioning whether it should be based on static or kinetic friction. Key equations discussed include the net forces acting on both blocks, with emphasis on the gravitational components along the slope. Clarifications are made regarding the role of gravity in the calculations, particularly its effect on the block on the incline. The conversation concludes with the user gaining a better understanding of the forces involved.
pb23me
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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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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:
 
well my only real question is what is the force due to friction? is it µ*mgcos\theta or µ*mgsin\theta
or is this equation correct ∑Fx=T-mg(cosθ)µ this is the equation for the block on the incline
 
Last edited:
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
 
pb23me said:
well my only real question is what is the force due to friction? is it µ*mgcos\theta or µ*mgsin\theta

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)
 
oh are you telling me that gravity is acting dwn and to the rght? i think i see what u mean
 
It's what gravity does best! :wink:

I'm off to bed now …

goodnight! :zzz:​
 
thanx for the help:)
 
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