Answer
349.2k+ views
Hint: Any block placed on a rough surface (having some coefficient of friction) moves when the force applied on it is greater than limiting static friction (or limiting friction) whose value is equals to the normal force times the coefficient of friction.
Formula used:
The mathematical expression for static friction is given by,
\[f\, = \,\mu N\]
where $f$ is the frictional force and $N$ is the normal force.
Complete step by step answer:
Let’s analyse the blocks first
Block C: Forces acting on Block C are:
-Normal reaction force from surface on which it is kept (N)
-Tension force (${T_1}$)
-Its weight (60N)
-Static frictional force (f)
As block C is in equilibrium
\[\therefore \,\,\sum {{F_x}\, = \,0\,} \]
\[ \Rightarrow \,f - \,{T_{1\,}}\, = \,0\]
Hence \[f = \,\,{T_{1\,}}\] ……………..…………………\[(i)\]
also \[\sum {{F_y}\, = \,0\,} \]
\[ \Rightarrow \,\,N\, - \,\,60\, = \,0\] or \[N\, = \,\,60\] …………..\[(ii)\]
The frictional force on block C:
\[f\, = \,\mu N\, = \,0.5 \times \,60\, = \,30N\]………\[(iii)\] [using \[(ii)\]]
Using equation \[(i)\] and \[(iii)\] we get \[{T_1}\,\, = \,\,30N\]………….\[(iv)\]
Block B: Forces acting on Block B are:
Tension force (${T_3}$)
Its weight (W). As block B is in equilibrium
\[\therefore \,\sum {{F_y}\, = \,0\,} \] \[ \Rightarrow \,\,{T_{3\,}}\, - \,\,W\, = \,0\]
Hence, \[{T_{3\,}}\, = \,\,W\] ………………………………..\[(v)\]
As the junction P is in equilibrium so at point P
\[\sum {{F_x}\, = \,0\,} \]
\[\therefore \,\,{T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}\]……………..\[(vi)\]
\[\sum {{F_y}\, = \,0\,} \]
\[\therefore \,{T_3}\, = \,\,{T_2}\, \times \dfrac{1}{{\sqrt 2 }}\]…………….\[(vii)\]
Now using equation \[(vi)\]and\[(iv)\]we get
\[{T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow \,{T_2} = \,\sqrt 2 {T_1}\,\]
\[\therefore \,\,{T_2}\,\, = \,30\sqrt 2 \,N\]……………\[(viii)\]
Finally, using equation \[(v)\],\[(vii)\] and \[(viii)\]
\[{T_3}\, = \,\,30\sqrt 2 \, \times \dfrac{1}{{\sqrt 2 }}\, = \,30N\] and \[{T_{3\,}}\, = \,\,W\]
Hence \[W\,\, = \,\,30\,Newtons\].
Note: If a block is at rest and keeps on resting on applying force then applied force is equal to the frictional force on the block as net force on the block = 0 in the condition of equilibrium. A body remains at rest if no external force is applied to it.
Formula used:
The mathematical expression for static friction is given by,
\[f\, = \,\mu N\]
where $f$ is the frictional force and $N$ is the normal force.
Complete step by step answer:
Let’s analyse the blocks first
![seo images](https://www.vedantu.com/question-sets/e780e957-d7ac-4c26-bc10-38ceb67581397049179544170924637.png)
Block C: Forces acting on Block C are:
-Normal reaction force from surface on which it is kept (N)
-Tension force (${T_1}$)
-Its weight (60N)
-Static frictional force (f)
![seo images](https://www.vedantu.com/question-sets/80a609ba-7624-431c-923f-4318a5c6fc084346941180276944090.png)
As block C is in equilibrium
\[\therefore \,\,\sum {{F_x}\, = \,0\,} \]
\[ \Rightarrow \,f - \,{T_{1\,}}\, = \,0\]
Hence \[f = \,\,{T_{1\,}}\] ……………..…………………\[(i)\]
also \[\sum {{F_y}\, = \,0\,} \]
\[ \Rightarrow \,\,N\, - \,\,60\, = \,0\] or \[N\, = \,\,60\] …………..\[(ii)\]
The frictional force on block C:
\[f\, = \,\mu N\, = \,0.5 \times \,60\, = \,30N\]………\[(iii)\] [using \[(ii)\]]
Using equation \[(i)\] and \[(iii)\] we get \[{T_1}\,\, = \,\,30N\]………….\[(iv)\]
Block B: Forces acting on Block B are:
Tension force (${T_3}$)
![seo images](https://www.vedantu.com/question-sets/db47080b-206c-4dc4-ae95-04812e5b4e679083058259798372432.png)
Its weight (W). As block B is in equilibrium
\[\therefore \,\sum {{F_y}\, = \,0\,} \] \[ \Rightarrow \,\,{T_{3\,}}\, - \,\,W\, = \,0\]
Hence, \[{T_{3\,}}\, = \,\,W\] ………………………………..\[(v)\]
As the junction P is in equilibrium so at point P
\[\sum {{F_x}\, = \,0\,} \]
\[\therefore \,\,{T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}\]……………..\[(vi)\]
\[\sum {{F_y}\, = \,0\,} \]
\[\therefore \,{T_3}\, = \,\,{T_2}\, \times \dfrac{1}{{\sqrt 2 }}\]…………….\[(vii)\]
Now using equation \[(vi)\]and\[(iv)\]we get
\[{T_1}\, = \,{T_2} \times \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow \,{T_2} = \,\sqrt 2 {T_1}\,\]
\[\therefore \,\,{T_2}\,\, = \,30\sqrt 2 \,N\]……………\[(viii)\]
Finally, using equation \[(v)\],\[(vii)\] and \[(viii)\]
\[{T_3}\, = \,\,30\sqrt 2 \, \times \dfrac{1}{{\sqrt 2 }}\, = \,30N\] and \[{T_{3\,}}\, = \,\,W\]
Hence \[W\,\, = \,\,30\,Newtons\].
Note: If a block is at rest and keeps on resting on applying force then applied force is equal to the frictional force on the block as net force on the block = 0 in the condition of equilibrium. A body remains at rest if no external force is applied to it.
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