A uniform ladder of length 5 m is placed against the wall as shown in the figure- If coefficient of friction mu is the same both the walls- what is the minimum value of mu it not to slip -mu -dfrac-1-4-mu -dfrac-1-2-mu -dfrac-1-3-mu -dfrac-1-5-

Question

A uniform ladder of length 5 m is placed against the wall as shown in the figure- If coefficient of friction mu  is the same both the walls- what is the minimum value of mu  it not to slip -mu -dfrac-1-4-mu -dfrac-1-2-mu -dfrac-1-3-mu -dfrac-1-5-

Toppr Admin · Accepted Answer

N1-x3BC-N2 -1-x3BC-N1-N2-mg-xA0-  -2-Torque about A- -x3C4-A-0-x21D2-3N2-x2212-4N1-x2212-52-xD7-35mg-0-xA0- -3-

A uniform ladder of length $5 m$ is placed against the wall as shown in the figure. If coefficient of friction $μ$ is the same for both the walls, what is the minimum value of $μ$ for it not to slip ?

$μ = \frac{1}{4}$
$μ = \frac{1}{2}$
$μ = \frac{1}{3}$
$μ = \frac{1}{5}$

$N_{1} = μ N_{2}$ ...(1)
$μ N_{1} + N_{2} = m g$ ...(2)

Torque about A: $τ_{A} = 0$
$\Rightarrow 3 N_{2} - 4 N_{1} - \frac{5}{2} \times \frac{3}{5} m g = 0$ ....(3)

A uniform ladder of length 5m is placed against the wall as shown in the figure. If coefficient of friction μ is the same for both the walls, what is the minimum value of μ for it not to slip ?μ=14μ=12μ=13μ=15

N1=μN2 ...(1)μN1+N2=mg ...(2)Torque about A: τA=0⇒3N2−4N1−52×35mg=0 ....(3)

A uniform ladder of length $5 m$ is placed against the wall as shown in the figure. If coefficient of friction $μ$ is the same for both the walls, what is the minimum value of $μ$ for it not to slip ?

$μ = \frac{1}{4}$
$μ = \frac{1}{2}$
$μ = \frac{1}{3}$
$μ = \frac{1}{5}$

$N_{1} = μ N_{2}$ ...(1)
$μ N_{1} + N_{2} = m g$ ...(2)

Torque about A: $τ_{A} = 0$
$\Rightarrow 3 N_{2} - 4 N_{1} - \frac{5}{2} \times \frac{3}{5} m g = 0$ ....(3)