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Calculating Friction Using Coeffcients

Two blocks m1...

Question

Two blocks $m_{1}$ and $m_{2}$ are resting on a rough inclined plane of inclination $37^{\circ}$ as shown in figure. The contact force between the two blocks is
(Given $m_{1} = 4 kg, m_{2} = 2 kg,$ $μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = 3 / 5$ )

A

3.2 N

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B

3.6 N

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C

7.2 N

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D

Zero

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Solution

The correct option is D Zero
For a mass sliding down on inclined plane with friction $μ$
$a = g sin θ - μ g cos θ$
As $μ_{1} > μ_{2}$ , $a_{1} < a_{2}$
i.e the acceleration of $m_{2}$ is greater than that of $m_{1}$ . Hence, there will not be any contact. i.e $N = 0$

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Similar questions

m_{1}

m_{2}

are resting on a rough inclined plane of inclination

37^{\circ}

as shown in figure. The contact force between the blocks is,

(m_{1} = 4 kg, m_{2} = 2 kg, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = 3 / 5)

m_{1}

m_{2}

are resting on a rough inclined plane of inclination

37^{\circ}

as shown in figure. The contact force between the blocks is

(m_{1} = 4 k g, m_{2} = 2 k g, μ_{1} = 0.8, μ_{2} = 0.5, g = 10 {m/s}^{2}, sin 37^{\circ} = \frac{3}{5})

m_{1} = 4 k g

m_{2} = 2 k g

, connected by a weightless rod on a plane having inclination of

37^{\circ}

as shown in figure. The coefficients of dynamic friction of

m_{1} a n d m_{2}

with the inclined plane are

μ = 0.25

. Then the common acceleration of the two blocks and the tension in the rod are (take sin37 = 3/5)

Q. In the figure shown the friction between the

4

kg block and the incline as

μ_{1}

8

kg and incline is

μ_{2}

g = 10 m / s^{2}

μ_{1} = 0.2

μ_{2} = 0.3

then find acceleration of

m_{1}

m_{2}

?

m_{1} = 4 k g

m_{2} = 2 k g

are connected by a weightess rod on a plane having inclination of

37^{\circ}

. The coefficient of dynamic friction for blocks

m_{1}

m_{2}

with the inclined plane is

μ = 0.2

μ_{k} = μ_{s} = 0.2

{m s}^{- 2}

. Then the common acceleration of the two blocks and the tension in the rod will be:

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