Determine the aceleration of points B,C and D, angular aceleration of rod and the tension in rope DE when ropes AB suddenly breaks

In order to solve the problem you need the following equations:

• equilibrium along the x-axis $$Tcostheta = m cdot a_{Cx} $$

• equilibrium along the y-axis $$Tsintheta - m g= m cdot a_{Cy} $$

• equilibrium about point C for bar BCD $$ vec{r}_{CD}times vec{T} = I alpha_{gamma, BD} $$

where:

• $a_{Cx},a_{Cy} $ the x and y components of acceleration for the middle of the bar
• $alpha_{gamma, BD}$ the angular acceleration of body BD
• $T$ is the magnitude of the force on the wire
• $r_{CD} = +5 vec{i}$ is the position vector from C to D

The above equations have 5 unknowns $(T, a_{Cx},a_{Cy} , alpha_{gamma, BD})$

You also have the following equation:

$$vec{a}_C = vec{a}_D + alpha_{gamma, BD}times vec{r}_{CD}$$

where:

• $vec{a}_D = alpha_{gamma, DE}times vec{r}_{ED}$
• $vec{r}_{ED} = -3vec{i} -4vec{j} =begin{bmatrix}-3-4end{bmatrix}$

so in matrix form the acceleration for point C can be written as

$$begin{bmatrix}a_{Cx}a_{Cy}end{bmatrix} = begin{bmatrix}0 alpha_{gamma,DE} end{bmatrix} times begin{bmatrix}-3-4end{bmatrix}+ begin{bmatrix}0 alpha_{gamma,BD} end{bmatrix} times begin{bmatrix}-5 end{bmatrix} $$

Now you have two additional equations and only one additional unkwown (angular acceleration of the wire $alpha_{gamma,DE}$). The total is 5 unkwowns, with 5 equations which can be solved.

I solved it for ($T,alpha_{gamma,DE},alpha_{gamma,BD}$ and the results are:

• $T= 134.3N$,
• $alpha_{gamma,DE}= 0.4031$
• $alpha_{gamma,BD}= 1.29$

if you then apply those values you get the acceleration at

• $a_B = 1.61260mathbf{vec{i}} - 14.110mathbf{vec{j}}$
• $a_C = 1.61260mathbf{vec{i}} - 7.659863$
• $a_D = 1.61260mathbf{vec{i}} - 1.2094520mathbf{vec{j}}$