Answer:
The minimum angle is 51.34°
Explanation:
Given that,
Weight of ladder = 100 N
Length = 8.0 m
Coefficient of static friction = 0.40
We need to calculate the normal force
Using Newtons law in vertical direction
[tex]F_{y}=n-mg[/tex]
[tex]N-mg=0[/tex]
[tex]N=mg[/tex]
We need to calculate the normal force
Using Newtons law in horizontal direction
[tex]F_{s}=f_{s}-P[/tex]
[tex]f_{s}-P=0[/tex]
[tex]f_{s}=P[/tex]
[tex]P=\mu mg[/tex]
We need to calculate the minimum angle
Using torque about the point A then
[tex]-mg\cos\theta\times AB+P\sin\theta\times AC=0[/tex]
Put the value into the formula
[tex]mg\cos\theta\times(\dfrac{L}{2})=\mu mg\sin\theta\times L[/tex]
[tex]\cos\theta\times\dfrac{1}{2}=\mu\sin\theta[/tex]
[tex]\dfrac{1}{2}=\mu\times\tan\theta[/tex]
[tex]\theta=\tan^{-1}(\dfrac{1}{2\times0.40})[/tex]
[tex]\theta=51.34^{\circ}[/tex]
Hence, The minimum angle is 51.34°
The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.
Given data:
The weight of ladder is, W = 100 N.
The length of ladder is, L = 8.0 m.
The coefficient of static friction between ladder and floor is, [tex]\mu =0.40[/tex].
Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,
[tex]N = mg[/tex]
And force along the horizontal direction is,
[tex]F= \mu \times N\\\\F = \mu \times mg[/tex]
Now, taking the torque along the either end of ladder as,
[tex]-mgcos \theta \times \dfrac{L}{2}+Fsin \theta \times L =0\\\\mgcos \theta \times \dfrac{L}{2} = Fsin \theta \times L[/tex]
Solving as,
[tex]mgcos \theta \times \dfrac{L}{2} = (\mu mg) \times sin \theta \times L\\\\cos \theta \times \dfrac{1}{2} = (\mu) \times sin \theta\\\\tan \theta = \dfrac{1}{2 \times 0.40 }\\\\\theta = tan^{-1}(1/0.80)\\\\\theta = 51.34^{\circ}[/tex]
Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.
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