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

$$ \text { The forces acting on the blocks can be described as follows: } $$$$\text { For } m_1 \text { : }$$$$\text { For } m_2 \text { : }$$$$\text { Normal Force: } N_1=m_1 g \cos \theta$$$$\text { Normal Force: } N_2=m_2 g \cos \theta$$$$\text { Friction Force: } f_1=\mu_1 N_1=\mu_1 m_1 g \cos \theta$$$$\text { Friction Force: } f_2=\mu_2 N_2=\mu_2 m_2 g \cos \theta$$$$\begin{aligned}&\text { Component of weight down the incline: }\\&m_1 g \sin \theta\end{aligned}$$$$\begin{aligned}&\text { Component of weight down the incline: }\\&m_2 g \sin \theta\end{aligned}$$$$\begin{aligned}&\text { Since } m_1 \text { is in equilibrium: }\\&\begin{gathered}m_1 g \sin \theta=T+f_1 \\T=m_1 g\left(\sin \theta-\mu_1 \cos \theta\right)\end{gathered}\end{aligned}$$$$\begin{aligned}&\text { Since } m_2 \text { is in equilibrium: }\\&\begin{gathered}m_2 g \sin \theta=T-f_2 \\T=m_2 g\left(\sin \theta+\mu_2 \cos \theta\right)\end{gathered}\end{aligned}$$Equating the two tensions:$$ m_1 g\left(\sin \theta-\mu_1 \cos \theta\right)=m_2 g\left(\sin \theta+\mu_2 \cos \theta\right) $$Cancelling out $g$ :$$ m_1\left(\sin \theta-\mu_1 \cos \theta\right)=m_2\left(\sin \theta+\mu_2 \cos \theta\right) $$The value of $\frac{m_1}{m_2}$ for the system to be in equilibrium is:$$ \frac{m_1}{m_2}=\frac{\mu_2 \cos \theta+\sin \theta}{\sin \theta-\mu_1 \cos \theta} $$

Laws Of Motion - IAT-IISER Physics PYQs

%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

Laws Of Motion

High-Yield Trend

Questions
1 MCQs

Official Solution

Laws Of Motion

High-Yield Trend

Questions 1 MCQs

Official Solution

Questions
1 MCQs