How Do You Calculate Mass M1 in a Frictional Inclined Plane Pulley System?

[twotaileddemon]

Why would the problem be wrong? Think about the essence of what's going on. There is a mass under full gravity pulling another mass being resisted by only partial gravity (i.e. $\mu g \cos \theta$). It would be an interesting exercise to calculate just how big M1 would need to be such that the acceleration is just barely zero...

You bring up a good point~

 

[bob1182006]

getting 10.1kg here

for m1 you have:

$$T-m_1g\sin\theta-f_k=m_1a$$
$$T-m_1g\sin\theta-\mu_km_1g\cos\theta=m_1a$$
Since Tension is pulling it up, gravity and friction are pulling down.

for m2:

$$T-m_2g=m_2a$$
$$T=m_2a+m_2g$$
since you know m2a (3.27 down) you can find a ~ -.503 m/s^2
and the tension ~60.495

plugging in the tension, gravity, angle, acceleration into the formula for m1 you can solve for m1 and get ~10.1kg

 

[hotvette]

Duh! I totally left out the component of gravity on M1! That would make M1 smaller, though, not bigger...

This is precisely what happens when one is rushing and isn't careful about including all the forces involved... geesh, and a homework helper should know better!

I'm now getting M1 < M2 like others originally thought. Not a good day...

 

[bob1182006]

wel if you consider the case where the angle is 0 and the block is just vertical like on the edge of a table. and you do the calculation for m1 again m1 will be like 21kg.

 

[FedEx]

See its a problem having no complications except the FBD

Here if you see at the system then you must can understand it all in a go.

The equations would be like this

$$m_2g - T = m_2a$$

$$T - m_1gsin\theta\ - {\mu\{m_1gcos}\theta\} = m_1a$$