Question

Block m1 still has a mass of 600 g and is on the same incline of 30 degrees. If the coefficient of friction is 0.1, what mass must block m2 have in order for it to accelerate downward at 1.5 m/s^2? *

Answer 1

@anb123

I think there something wrong with the problem. As far as I understand the acceleration doesn't depend on the mass of the object, but just on the coefficient of friction and the angle of the incline. Take a look at this (I'll consider 0.1 as the coefficient of kinetic and static friction):

\[F_{net} = F_{exerted} - f_{k}\]

Fnet is the acceleration of 1.5m/s² of the box with m₂: (1.5m/s²)(m₂)
Fexerted this is the componet of the weight of the box that is causing the slip: (m₂g)(sinθ)
Fk is the kinectic friction = μN = μ(m₂g)(cosθ)

\[(1.5\ m/s^2)m_2= (m_2g)\sin(\theta) - (m_2g)\cos(\theta)\]

Divide both sides by m₂ and you get:

\[1.5\ m/s^2= g \sin(\theta) - \mu_k g \cos(\theta)\]
\[1.5\ m/s^2= g (\sin(\theta) - \mu_k \cos(\theta))\]
\[1.5\ m/s^2= 9.8m/s^2 (\sin(30º) - 0.1 \cos(30º))\]
\[1.5\ m/s^2= 4.1\ m/s^2\]

The equation doesn't hold.

If the box 1 doesn't slide, then the static friction is equal to the component of the weight of the box that would causing it to slip:

\[F_{exerted} = f_{s}\]
\[m_1g\sin(\theta) = \mu_sm_1g\cos(\theta)\]
\[\sin(\theta) = \mu_s\cos(\theta)\]
\[\frac{\sin(\theta)}{\cos(\theta)} = \mu_s\]
\[\tan(\theta) = \mu_s \]
\[\theta = \tan^{-1} (\mu_s) \]
\[\theta = \tan^{-1} (0.1) \]
\[\theta = 5.7º \]

That would have to be angle of the incline in other for m₁ remain at rest. In both cases mass cancels (as it must be).

Maybe I've misunderstood something.