Minimum mass M to pull block up incline with spring

[chacha1234]

Homework Statement

1] A block of mass 'm' is attached with a massless spring of force constant 'k'. The block is placed over a rough inclined surface for which the coefficient of friction is 3/4. The minimum value of 'M' required to move the block up the place is ...

[A] 3mg/5 6mg/5 [C] 2mg [D] mg/5.
http://img26.imageshack.us/img26/1318/72640151.jpg

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*[Note: sin37 = 3/5, cos37 = 4/5, tan37 = 3/4].

* this note is NOT given in the problem.

please solve this for me.

Homework Equations

The Attempt at a Solution

 

[Hootenanny]

please solve this for me.

No, we will not solve this for you, but we will help you solve it.

 

[chacha1234]

From the figure . kx=T.
Forming the different equations, we have: Mg - T = Ma
Since the block just has to move up, i.e., it just has to start its motion, so we'll consider the acceleration to be zero.
So, now we have: Mg = T

Similarly, we'll get the other equation as
T - mg.Sin37 - (mu).mg.Cos37 = 0 , [mu = coefficient of friction=3/4]
T=Mg . the answer as: M = 6m/5
http://img190.imageshack.us/img190/1653/figurek.jpg

but the answer given is : 3m/5.

this problem is given at four locations and all with the same answer: 3m/5

 

[CFDFEAGURU]

I see some fault with the FBD. The normal force on the sliding block is in the wrong direction and you are missing the downward gravitational force for the sliding block.

Thanks
Matt

 

[Hootenanny]

Your free body diagram looks fine to me, but I can't seem to get 3/5m as the answer. I've called in some of the other Homework Helpers for advice.

By the way, these:

[Note: sin37 = 3/5, cos37 = 4/5, tan37 = 3/4].

are not correct.

 

[CFDFEAGURU]

Ohh sorry, The arrow through me off. The normal force is in the correct direction. I show the arrow pointing at the bottom of the box, which is equivalent to what you have shown. It just didn't click.

Thanks
Matt

 

[Fightfish]

Not very sure about this, but if you consider the block m alone, it is able to stay in equilibrium on the inclined surface purely due to friction.
Thus to instantaneously move it, we would just need to overcome the static friction opposing its motion up the plane, which is equal to 3mg/5.

 

[D H]

this problem is given at four locations and all with the same answer: 3m/5

Can you give some references, please?

 

[dx]

The answer 3m/5 is correct. Here's a hint: Assume the friction coefficient is infinite. Before the block of mass m starts to move, the spring will stretch. How much can the gravitational force on M stretch the spring before the reaction force of the spring on the block M becomes equal to the gravitational force and the system reaches equilibrium?

 

[D H]

I disagree. There is no force doubling through the spring.

Edit
Everything else I said is wrong.

 

[chacha1234]

Thanks to all of u.
CFDFEAGURU: pls have a second look at the diagram.

Hootenanny:

Note: sin37 = 3/5, cos37 = 4/5, tan37 = 3/4].
are not correct.

These values are approx. values, else we cannot meet the answer in the options given therein.

Fightfish & dx: Can u pls post a diagram of what u said to explain the answer (3m/5)? bcoz the block is about to move up the incline (for minimum of M), so gravity componenet and friction both r in the downward direction.

D H:
one of my friends gave me this, source ... may be from mechanics of arihant, chapter could be: laws of motion including friction.

 

[D H]

Depending on the assumptions made regarding the initial state of the mass M we Homework Helpers (Hootenay talked about this in post #5) arrived at either 6/5 m or 3/5 m. A couple of questions to help clarify:

- Is the question stated in the original post exact, or did you leave some information out?
- Is the question from a section of the book on statics or dynamics?

 

[chacha1234]

thanks all i finally find the solution

 

[Dweirdo]

Ahm,
if M will be 3/5 m, it still won't be able to move m(friction won't let it move, although at that point its close to zero,and the moment it goes kinetic, boom...).
IMO.