Answer:
A) 0.188
B) 53.1 ⁰
Explanation:
taking moment about 0
∑ Mo = Lo∝
mg 1/2 sin∅ = 1/3 m L^2∝
note ∝ = w[tex]\frac{dw}{d}[/tex]∅
forces acting along t-direction ( ASSUMED t direction)
∑ Ft = Ma(t) = mr∝
mg sin ∅ - F = m* 1/2 * 3g/2l sin∅
therefore F = mg/4 sin∅
forces acting along n - direction ( ASSUMED n direction)
∑ Fn = ma(n) = mr([tex]w^{2}[/tex])
= mg cos∅ - N = m*1/2*3g/1 ( 1 - cos∅ )
hence N = mg/2 ( 5cos∅ -3 )
A ) Angle given = 30⁰c find coefficient of static friction
∪ = F/N
= [tex]\frac{\frac{mg}{4}sin30 }{\frac{mg}{2}(5cos30 -3) }[/tex] = 0.188
B) when there is no slip
N = O
= 5 cos ∅ -3 =0
therefore cos ∅ = 3/5 hence ∅ = 53.1⁰
