Answer
427.8k+ views
Hint: To solve the above problem we will have to prepare a free body diagram to understand all forces that are acting on the body. After that we will balance the forces which are acting in the opposite direction then we will find the unknown quantity that is asked in the question.
Complete answer:
So, the free body diagram showing all the forces on the man and the plank is as follows,
So, from the free body diagram we see that the man is experiencing three forces and balancing the forces in the opposite direction we will have the following equation,
$T + R = Mg$
$ \Rightarrow R = Mg - T$-----equation (1)
Now on seeing the free body diagram of the plank, balancing the vertical forces,
In vertical upward direction the normal reaction force ${R_1}$ and in the vertical downward direction mass of the plank$mg$and the other normal reaction force $R$is acting.
${R_1} = mg + R$
$ \Rightarrow {R_1} = R + mg$--------equation (2)
In the horizontal direction the force on the left horizontal is friction force $\mu {R_1}$and the force in the right horizontal direction is the tension in the rope $T$.
So, $T = \mu {R_1}$------equation (3)
Now, Putting the value of the ${R_1}$from the equation (2), we get
$ \Rightarrow T = \mu (R + mg)$
$ \Rightarrow T = \mu R + \mu mg$
$ \Rightarrow T = \mu (Mg - T) + \mu mg$
$ \Rightarrow T = \mu Mg - \mu T + \mu mg$
$ \Rightarrow T + \mu T = \mu Mg + \mu mg$
$ \Rightarrow T(1 + \mu ) = \mu (Mg + mg)$
$ \Rightarrow T = \dfrac{{\mu (Mg + mg)}}{{(1 + \mu )}}$
Hence, the maximum force that the man can exert on the rope is $\dfrac{{\mu (Mg + mg)}}{{(1 + \mu )}}$.
So, the correct answer is “Option D”.
Note:
Here we have taken the help of the free body diagram so we should know the appropriate definition of the free body diagram. Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. Sometimes it is called a force diagram since it shows the proper position and direction of the forces.
Complete answer:
So, the free body diagram showing all the forces on the man and the plank is as follows,
![seo images](https://www.vedantu.com/question-sets/9ef9bce5-ceba-407a-a7f7-a62dfb298ddc7109969216332656900.png)
So, from the free body diagram we see that the man is experiencing three forces and balancing the forces in the opposite direction we will have the following equation,
$T + R = Mg$
$ \Rightarrow R = Mg - T$-----equation (1)
Now on seeing the free body diagram of the plank, balancing the vertical forces,
In vertical upward direction the normal reaction force ${R_1}$ and in the vertical downward direction mass of the plank$mg$and the other normal reaction force $R$is acting.
${R_1} = mg + R$
$ \Rightarrow {R_1} = R + mg$--------equation (2)
In the horizontal direction the force on the left horizontal is friction force $\mu {R_1}$and the force in the right horizontal direction is the tension in the rope $T$.
So, $T = \mu {R_1}$------equation (3)
Now, Putting the value of the ${R_1}$from the equation (2), we get
$ \Rightarrow T = \mu (R + mg)$
$ \Rightarrow T = \mu R + \mu mg$
$ \Rightarrow T = \mu (Mg - T) + \mu mg$
$ \Rightarrow T = \mu Mg - \mu T + \mu mg$
$ \Rightarrow T + \mu T = \mu Mg + \mu mg$
$ \Rightarrow T(1 + \mu ) = \mu (Mg + mg)$
$ \Rightarrow T = \dfrac{{\mu (Mg + mg)}}{{(1 + \mu )}}$
Hence, the maximum force that the man can exert on the rope is $\dfrac{{\mu (Mg + mg)}}{{(1 + \mu )}}$.
So, the correct answer is “Option D”.
Note:
Here we have taken the help of the free body diagram so we should know the appropriate definition of the free body diagram. Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation. Sometimes it is called a force diagram since it shows the proper position and direction of the forces.
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