Find the Acceleration of a swinging bar and the forces on the struts

Question

https://prnt.sc/G96geXdw_COL

The link leads to the picture of the problem. Essentially the problem is a 12 kg bar (4m in length) attached to the ceiling by 2 massless struts held initially at an angle of 50 degrees with the horizontal. One strut is on the left edge of the beam which the other is 2.5 m from the left edge. The problem asks to find the acceleration and forces on the struts when the bar is let go. The bar will swing down and to the left.

Answer 1

Let

m = 12 kg be the mass of the beam,

r = 2 m be the length of the two struts,

θ = 50° be the initial angle of the struts,

d1 = 2 m be the distance between point B and the beam's center of gravity,

d2 = 0.5 m be the distance between point C and the beam's center of gravity,

T1 (unknown) be the tension force in the left strut,

T2 (unknown) be the tension force in the right strut,

g = 9.81 m/s² be gravitational acceleration,

v(t) be the velocity of the beam at time t.

We are given v(0) = 0.

Once the beam starts to move, its center of gravity follows a circular

trajectory with radius r around a point between A and D.

Notice that the structs, being 2-force elements, experience tension parallel to their lengths.

Besides these two tensions, the only other force acting on the beam is gravity.

Let's decompose the forces into two components --

radial, i.e. in the direction of the struts, and

tangential, i.e., perpendicular to the struts.

The structs have 0 projection on the tangential direction.

They are both directed radially towards the upper left.

The force of gravity has a radial component mgsin(θ), pointing towards the lower right, and

tangential component mgcos(θ) pointing towards the lower left.

Thus the net radial force, directed towards the upper left, is

T1 + T2 - mgsin(θ)

And the net tangential force, directed towards the lower left, is

mgcos(θ)

Let a_r be the radial component of acceleration, and a_t be the tangential component.

By Newton's second law

T1 + T2 - mgsin(θ) = ma_r (1)

mgcos(θ) = ma_t (2)

From (2)

a_t = gcos(θ)

By properties of rotational motion

a_r(t) = v²(t)/r

At time 0, v(0) = 0, and therefore

a_r(0) = 0

Therefore at time 0 equation (1) becomes

T1 + T2 - mgsin(θ) = 0 (3)

The beam is not rotating around its center of gravity, therefore the net torque around the center must be 0:

T1d1sin(θ) = T1d2sin(θ)

After simplification

T1d1 = T2d2 (4)

Solving equations (3) and (4)

T1 = d2mgsin(θ)/(d1 + d2)

T2 = d1mgsin(θ)/(d1 + d2)

Substituting actual numbers:

a_r= 0

a_t = 9.81×cos(50°) = 6.3 m/s²

T1 = (0.5×12×9.81×sin(50°))/2.5 = 18 N

T2 = (2×12×9.81×sin(50°))/2.5 = 72 N