Question

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.

Answer 1

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

Step-by-step 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

`-mg((lcos\theta)/(2))=(1)/(3)(ml^(2))\alpha`

`\alpha=(-3gcos\theta)/(2l)`

Let G be the center of gravity of the bar AB

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

`\vec{r_(G)}=[(lcos\theta)/(2)\hat{i}-(lcos\theta)/(2)\hat{j}]`

The acceleration at the center of the bar

`\vec{a_(G)}=\vec{a_(a)}+\vec{\alpha}X\vec{r_(G)}-\omega^(2)\vec{r_(G)}`

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

`(\vec{a_(G)})_(x)\hat{i}+(\vec{a_(G)})_(y)\hat{j}=0+((-3gcos\theta)/(2l))\hat{k}*[(lcos\theta)/(2)\hat{i}-(lcos\theta)/(2)\hat{j}]-\omega^(2)[(lcos\theta)/(2)\hat{i}-(lcos\theta)/(2)\hat{j}]`

`(\vec{a_(G)})_(x)\hat{i}+(\vec{a_(G)})_(y)\hat{j}=[-(cos\theta(2l\omega^(2)+3gsin\theta))/(4)\hat{i}+((2l\omega^(2)sin\theta-3gcos^(2)\theta)/(4))\hat{j}]`

Comparing the coefficients of i

`=-(cos\theta(2l\omega^(2)+3gsin\theta))/(4)`

Comparing coefficients of j

`(\vec{a_(G)})_(y)=(2l\omega^(2)sin\theta-3gcos^(2)\theta)/(4)`

Net force on x direction

`F_(x)=(\vec{a_(G)})_(x)`

substituting the values

`F_(x)`=1.5(14.58L+11.96)

Similarly net force on y direction

`F_(y)=(\vec{a_(G)})_(y)+mg`

= 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

`F=\sqrt{F_(x)^(2)+F_(y)^(2)}`

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