Question

End A of the uniform 5-kg bar is pinned freely to the collar, which has an acceleration = 4 m/s2 along the fixed horizontal shaft. If the bar has a clockwise angular velocity ω = 2 rad/s as it swings past the vertical, determine the components of the force on the bar at A for this instant.

Answer 1

Fa = 57.32 N

Step-by-step explanation:

given data

mass = 5 kg

acceleration = 4 m/s²

angular velocity ω = 2 rad/s

solution

first we take here moment about point A that is

∑Ma = Iα + ∑Mad ...............1

put here value and we get

so here I = (
`(1)/(12)` ) × m × L² ................2

I = (
`(1)/(12)` ) × 5 × 0.8²

I = 0.267 kg-m²

and

a is = r × α

a = 0.4 α

so now put here value in equation is 1

0 = 0.267 α + m r α (0.4) - m A (0.4)

0 = 0.267 α + 5 (0.4α) (0.4 ) - 5 (4) 0.4

so angular acceleration α = 7.5 rad/s²

so here force acting on x axis will be

∑ F(x) = m a(x) ..............3

a(x) = m a - m rα

put here value

a(x) = 5 × 4 - 5 × 0.4 × 7.5

a(x) = 5 N

and

force acting on y axis will be

∑ F(y) = m a(y) .............. 4

a(y) - mg = mrω²

a(y) - 5 × 9.81 = 5 × 0.4 × 2²

a(y) = 57.1 N

so

total force at A will be

Fa =
`√(a(x)^2+a(y)^2)` ...............5

Fa =
`√((57.1)^2+(5)^2)`

Fa = 57.32 N