Question

Three blocks of mass $10\text{kg}$, $5\text{kg}$ and $8\text{kg}$ are connected to each other with a light string and placed on a wedge with angle of inclination ${30}^{\circ }$. A force F applied on the string causes all the three blocks to move with a common acceleration of $2m/s^2$. If the coefficient of friction between the blocks and the surface of the wedge is $0.2$, then what is the tension at the three points $A,B$ and $C$?

• A. $T_A=200.8N,T_B=69.85N,T_C=113.51N$
• B. $T_A=113.51N,T_B=69.85N,T_C=200.8N$
• C. $T_A=200.8N,T_B=113.51N,T_C=69.85N$
• D. $T_A=195N,T_B=60N,T_C=100N$

Answer 1

The correct option is C $T_A=200.8N,T_B=113.51N,T_C=69.85N$

FBD of the three blocks taken as a system:



FBD of block C:



FBD of block B and C taken as a system:


The entire system accelerates at the rate of $2{\text{m/s}}^2$. Considering all the three masses together, we get,

$F-\mu (m_1+m_2+m_3)g\cos \theta -(m_1+m_2+m_3)g\sin \theta =(m_1+m_2+m_3)a\Rightarrow F-0.2\times (10+5+8)\times 10\times \cos {30}^{\circ }-(10+5+8)\times 10\times \sin {30}^{\circ }=(10+5+8)\times 2\Rightarrow F=200.8N$
Therefore tension at point A :$T_A=200.8N$

Due to the force exerted at point $B$, the other two blocks move with an acceleration of $2m/s^2$.
From the free body diagram, we get,
$F_B-0.2(5+8)\times 10\times \cos {30}^{\circ }-(5+8)\times 10\times \sin {30}^{\circ }=(5+8)\times 2\Rightarrow F_B=113.51N$
Therefore tension at point $B$ :$T_B=113.51N$

Due to the force exerted at point $C$, the last block moves with an acceleration of $2m/s^2$

From the free body diagram, we get,
$F_C-0.2\times 8\times 10\times \cos {30}^{\circ }-8\times 10\times \sin {30}^{\circ }=8\times 2\Rightarrow F_C=69.85N$
Therefore tension at point $C$ :$T_C=69.85N$