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

Answer:

Fa = 57.32 N

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 = ( $\frac{1}{12}$ ) × m × L²    ................2

I = ( $\frac{1}{12}$ ) × 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 = $\sqrt{a(x)^2+a(y)^2}$    ...............5

Fa = $\sqrt{(57.1)^2+(5)^2}$  

Fa = 57.32 N