The uniform slender bar ab has a mass of 6.4 kg and swings in a vertical plane about the pivot at a. if θ˙ = 2.7 rad/s when θ = 24°, compute the magnitude of the force supported by the pin at a at that instant.

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar,  θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

                        ∑ Mₐ = Iα

Where,     Mₐ - the moment about A

                 α  - angular acceleration

                 I - moment of inertia of the rod AB

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

The acceleration at the center of the bar

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

Comparing the coefficients of i

Comparing coefficients of j

Net force on x direction

substituting the values

=1.5(14.58L+11.96)

Similarly net force on y direction

               = 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

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