A uniform rod is made to lean between rough vertical wall and the ground. Friction coefficient between rod and wall is μ1=12 and between the rod and the ground is μ2=14. The rod is about to slip at contact surfaces. The correct options are:

For equilibrium of rod ∑Fx=0; N1=μ2N2 ..............(1) ∑FY=0⇒μ1N1+N2=W.......(2) Solving, N1=μ2W1+μ1μ2So, choice (1) is correct and N2=W1+μ1μ2So, choice (2) is correct, by substituting the values N1=2w9 & N2=8w9 ∴N2>N1

A uniform rod is made to lean between rough vertical wall and the ground. Friction coefficient between rod and wall is μ1=12 and between the rod and the ground is μ2=14. The rod is about to slip at contact surfaces. The correct options are:

For equilibrium of rod ∑Fx=0; N1=μ2N2 ..............(1) ∑FY=0⇒μ1N1+N2=W.......(2) Solving, N1=μ2W1+μ1μ2So, choice (1) is correct and N2=W1+μ1μ2So, choice (2) is correct, by substituting the values N1=2w9 & N2=8w9 ∴N2>N1

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A uniform rod is made to lean between rough vertical wall and the ground. Friction coefficient between rod and wall is $μ_{1} = \frac{1}{2}$ and between the rod and the ground is $μ_{2} = \frac{1}{4}$ . The rod is about to slip at contact surfaces. The correct options are:

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The normal reaction between rod and wall is $\frac{μ_{2} W}{1 + μ_{1} μ_{2}}$

$N_{1} > N_{2}$

$N_{2} > N_{1}$

The normal reaction between rod and ground is $\frac{W}{1 + μ_{1} μ_{2}}$

answer is A, B, C.

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Detailed Solution

For equilibrium of rod

$\sum F_{x} = 0; N_{1} = μ_{2} N_{2} . . . . . . . . . . . . . . (1) \sum F_{Y} = 0 \Rightarrow μ_{1} N_{1} + N_{2} = W . . . . . . . (2) S o l v i n g, N_{1} = \frac{μ_{2} W}{1 + μ_{1} μ_{2}}$