What Is the Minimum Mass M1 to Prevent Slipping on a 19.1° Slope?

[cclawfjj]

A block of mass M resting on a 19.1° slope is shown. The block has coefficients of friction μs=0.788 and μk=0.482 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 1.86 kg. What is the minimum mass M1 that will stick and not slip?

http://capa.physics.mcmaster.ca/figures/kn/Graph08/kn-pic0836.png


In terms of mass 1, following eqn can be written:

m1g sin 19.1 - T - Ff = 0

which can be rewritten as:

m1g sin 19.1 - T - m1g cos 19.1 (friction coefficient) = 0
right? since the object is at rest.
then i found T by equating it to m2g (T = m2g...see the diagram as to why i have donet this)

Hence m1g sin 19.1 - m2g - m1g cos 19.1 (0.788) = 0
and then solved for m1...but I am not getting the right answer
I don't understand why my approach is wrong...where did i go wrong here?

 

[chanvincent]

Here is the rule in this forums...

https://www.physicsforums.com/showthread.php?t=94379

 

[cclawfjj]

please help, much appreciated

 

[cclawfjj]

Plx

im literally bursting into tears right now, please
this question has been frustrating me for the last couple of days

 

[cclawfjj]

ppl don't be so cruel, please!11

 

[cclawfjj]

I've posted my solution and clearly demonstrated my attempt to resolve this question to no avail.

PLZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZZ
i beg you, you can have my undying gratitude

 

[chanvincent]

m1g sin 19.1 - T - Ff = 0

Apparently, you mess up the sign in Ff. When you are pulling the mass 1 uphill with tension T, the friction force shall point to the opposite to T.