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A boy with a mass of 4 kg is standing on a piece of wood with a mass of 5 kg. If the coefficient of friction between the wood and the floor is 0.5, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is ________N. (Round off to the Nearest Integer)
[Take \[g = 10{\text{ m/}}{{\text{s}}^2}\]]


Answer
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162.9k+ views
Hint: In this question, we need to determine the maximum force that the boy can exert. For this, we need to equate the frictional force (T) and the force exerted by the body. Also, we will use the following formula to get the desired result.

Formula used:
The formula for frictional force is given by
\[F = \mu N\]
Where, \[F\] is the frictional force, \[\mu \] is the coefficient of friction and \[N\] is the normal force.

Complete step by step solution:
We can say that the frictional force is equal to the force exerted by a boy.
\[F = T\]
But we know that\[F = \mu N\]
So, we get
\[\mu N = T\]

According to the given figure, we can say that \[N + T = 90\]
Also, \[\mu = 0.5\]
Thus, \[N = 90 - T\]

By putting these values in equation \[\mu N = T\]
So, we get
\[0.5\left( {90 - T} \right) = T\]
\[\Rightarrow 45 - 0.5T = T\]
\[\Rightarrow T + \dfrac{T}{2} = 45\]
By multiplying 2 on both sides, we get
\[2T + T = 90\]
\[\Rightarrow 3T = 90\]
By simplifying, we get
\[\therefore T = 30\]N

Therefore, the maximum force that the boy can exert on the rope so that the piece of wood does not move from its place is 30 N.

Note: Many students make mistakes in writing the formula for the frictional force. As a result, the end result may get wrong. Here, the simplification part for calculating normal force is also important for getting the final answer.