Laws Of Motion

Consider a mass-pulley system as shown in the figure. There is a wedge of mass

M

and equal wedge angles

%5Ctheta

lying on a rigid horizontal table. The coefficient of friction between the wedge and the table is

%5Cmu

. There are two blocks of mass

m_1

and

m_2

lying on the incline of the wedge. The coefficients of friction between the blocks and wedge are

%5Cmu_1

and

%5Cmu_2

as shown in the figure. Consider

m_1%3Em_2

and the coefficients of friction (

%5Cmu%2C%20%5Cmu_1

and

%5Cmu_2

) to be less than

%5Ctan%20%5Ctheta

. Gravity is acting downwards with acceleration due to gravity

g

. What should be the value of

%5Cfrac%7Bm_1%7D%7Bm_2%7D

so that the system is in equilibrium?

%5Ctext%20%7B%20The%20forces%20acting%20on%20the%20blocks%20can%20be%20described%20as%20follows%3A%20%7D

%5Ctext%20%7B%20For%20%7D%20m_1%20%5Ctext%20%7B%20%3A%20%7D

%5Ctext%20%7B%20For%20%7D%20m_2%20%5Ctext%20%7B%20%3A%20%7D

%5Ctext%20%7B%20Normal%20Force%3A%20%7D%20N_1%3Dm_1%20g%20%5Ccos%20%5Ctheta

%5Ctext%20%7B%20Normal%20Force%3A%20%7D%20N_2%3Dm_2%20g%20%5Ccos%20%5Ctheta

%5Ctext%20%7B%20Friction%20Force%3A%20%7D%20f_1%3D%5Cmu_1%20N_1%3D%5Cmu_1%20m_1%20g%20%5Ccos%20%5Ctheta

%5Ctext%20%7B%20Friction%20Force%3A%20%7D%20f_2%3D%5Cmu_2%20N_2%3D%5Cmu_2%20m_2%20g%20%5Ccos%20%5Ctheta

%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Component%20of%20weight%20down%20the%20incline%3A%20%7D%5C%5C%26m_1%20g%20%5Csin%20%5Ctheta%5Cend%7Baligned%7D

%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Component%20of%20weight%20down%20the%20incline%3A%20%7D%5C%5C%26m_2%20g%20%5Csin%20%5Ctheta%5Cend%7Baligned%7D

%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Since%20%7D%20m_1%20%5Ctext%20%7B%20is%20in%20equilibrium%3A%20%7D%5C%5C%26%5Cbegin%7Bgathered%7Dm_1%20g%20%5Csin%20%5Ctheta%3DT%2Bf_1%20%5C%5CT%3Dm_1%20g%5Cleft(%5Csin%20%5Ctheta-%5Cmu_1%20%5Ccos%20%5Ctheta%5Cright)%5Cend%7Bgathered%7D%5Cend%7Baligned%7D

%5Cbegin%7Baligned%7D%26%5Ctext%20%7B%20Since%20%7D%20m_2%20%5Ctext%20%7B%20is%20in%20equilibrium%3A%20%7D%5C%5C%26%5Cbegin%7Bgathered%7Dm_2%20g%20%5Csin%20%5Ctheta%3DT-f_2%20%5C%5CT%3Dm_2%20g%5Cleft(%5Csin%20%5Ctheta%2B%5Cmu_2%20%5Ccos%20%5Ctheta%5Cright)%5Cend%7Bgathered%7D%5Cend%7Baligned%7D

Equating%20the%20two%20tensions%3A

m_1%20g%5Cleft(%5Csin%20%5Ctheta-%5Cmu_1%20%5Ccos%20%5Ctheta%5Cright)%3Dm_2%20g%5Cleft(%5Csin%20%5Ctheta%2B%5Cmu_2%20%5Ccos%20%5Ctheta%5Cright)

Cancelling%20out

%3A

m_1%5Cleft(%5Csin%20%5Ctheta-%5Cmu_1%20%5Ccos%20%5Ctheta%5Cright)%3Dm_2%5Cleft(%5Csin%20%5Ctheta%2B%5Cmu_2%20%5Ccos%20%5Ctheta%5Cright)

The%20value%20of

\frac{m_1}{m_2}

for%20the%20system%20to%20be%20in%20equilibrium%20is%3A

%5Cfrac%7Bm_1%7D%7Bm_2%7D%3D%5Cfrac%7B%5Cmu_2%20%5Ccos%20%5Ctheta%2B%5Csin%20%5Ctheta%7D%7B%5Csin%20%5Ctheta-%5Cmu_1%20%5Ccos%20%5Ctheta%7D

Chapter Questions
1 MCQs