Acceleration of mass with friction

[Jonski]

Homework Statement

At the instant shown the block is moving down the slope with P= 49 N, β= 38°, θ= 33° and M= 33 kg.

Homework Equations

What is the acceleration of the block up the slope if the kinetic coefficient of friction is 0.5?

The Attempt at a Solution

So I split all the forces into the normal direction and parallel to the plane

Forces in the normal = 0 = N - mgcos33 + 49sin38
N = 241.34

Forces parallel to the plane = 49cos38 - mgsin33 + 0.5N
= -17.036
F/m = a
a = -17.036/33 = -0.516
But I am not getting the right answer, please help

 

[Qwertywerty]

Draw the FBD for mass M.

 

[Jonski]

 

[Jonski]

Draw the FBD for mass M.

Now what, I added up all the forces in each direction. Not sure where I went wrong

 

[Qwertywerty]

Oh, my earlier post didn't load.

First of, does friction oppose or support the motion of the block ?
And second, is the force on M being exerted by P, or the pulley ?

Hope this helps.

 

[haruspex]

Forces parallel to the plane = 49cos38 - mgsin33 + 0.5N

You are missing a force. Consider the forces acting on the pulley.
(I strongly encourage you to work entirely symbolically, only plugging in numbers right at the end. It has many advantages. See section 1 of https://www.physicsforums.com/insights/frequently-made-errors-equation-handling/)

 

[haruspex]

does friction oppose or support the motion of the block ?

We are told that acceleration is up the slope but the present velocity is down.

 

[Jonski]

You are missing a force. Consider the forces acting on the pulley.

I think the force I am missing is tension in the pulley,
Not sure but is it mgsin(theta)

 

[haruspex]

I think the force I am missing is tension in the pulley,
Not sure but is it mgsin(theta)

Not sure what you mean by tension "in" the pulley. The rope runs around the pulley. It will be more convenient to think of it as two separate ropes, but you need know the tension in each (which will be?)

 

[Jonski]

It will be more convenient to think of it as two separate ropes, but you need know the tension in each (which will be?)

Would it be p + mgsin(theta)

 

[haruspex]

Would it be p + mgsin(theta)

What is the tension in the upper part of the rope?

 

[Jonski]

What is the tension in the upper part of the rope?

Is it 49N ?

 

[Qwertywerty]

We are told that acceleration is up the slope but the present velocity is down.

Oops ! Missed the down.

 

[Jonski]

Oops ! Missed the down.

so P + mgsin(theta) + mgcos(theta)*friction

 

[n_freitass]

Sorry my bad..

 

[haruspex]

Is it 49N ?

Yes, it's P, which is given as 49N.
It doesn't say, but you should take the pulley as massless and having no axle friction. If the tension in the upper part of the rope is P, what must it be in the lower part? Hint: think about the net torque on the pulley.

 

[haruspex]

Would your force parallel be $F_p : \mu_kN + P + Pcos(\beta) - mg sin(\theta) = ma_p$ ?
And force normal be $F_n : Psin(\beta)+N-mgcos(\theta) = 0$

Please do not provide answers. The system on the homework forums is that we provide hints, explain misconceptions, point out algebraic errrors, etc.