
A wedge of mass $2m$ and a cube of mass m are shown in figure. Between the cube and wedge, there is no friction. The minimum coefficient of friction between wedge and ground so that wedge does not move is

(A) $0.10$
(B) $0.20$
(C) $0.25$
(D) $0.50$
Answer
233.4k+ views
Hint In these kinds of situations the first thing important is to notice the given problem statement. We can observe that this is a question for the straight approach. In this question we will have to simplify the equation to obtain all the known quantities for the final equation.
Complete step by step answer:
Here, first we have to sort out the given quantities:
Mass of the wedge: $2m$
Mass of the cube:$1m$
And the angle between surface and slope: $\theta = {45^o}$
Now, the equation balancing the forces:
$
\lambda \sin \theta = Fr \\
\lambda \sin \theta = \mu (2mg + mg{\cos ^2}\theta ) \\
mg\sin \theta \cos \theta = \mu (2mg + mg{\cos ^2}\theta ) \\
$
So, after interpreting the final equation with all know quantities we just have to put the values:
$
\dfrac{1}{2} = \mu (2 + \dfrac{1}{2}) \\
\mu = \dfrac{1}{5} \\
$
Note Most important thing here is to notice the given values, then we can easily interpret the equations. In these types of questions using the correct method is necessary unless it will get twisted.
Complete step by step answer:
Here, first we have to sort out the given quantities:
Mass of the wedge: $2m$
Mass of the cube:$1m$
And the angle between surface and slope: $\theta = {45^o}$
Now, the equation balancing the forces:
$
\lambda \sin \theta = Fr \\
\lambda \sin \theta = \mu (2mg + mg{\cos ^2}\theta ) \\
mg\sin \theta \cos \theta = \mu (2mg + mg{\cos ^2}\theta ) \\
$
So, after interpreting the final equation with all know quantities we just have to put the values:
$
\dfrac{1}{2} = \mu (2 + \dfrac{1}{2}) \\
\mu = \dfrac{1}{5} \\
$
Note Most important thing here is to notice the given values, then we can easily interpret the equations. In these types of questions using the correct method is necessary unless it will get twisted.
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