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

Question

The solutions are F(DE)=134.38N, a(D)=2.016m/s^2 , α=1.29rad/s^2, a(C)=7.828m/s^2

I agree that the FBD and the KD are these:

Then to solve this problem I've done:

$$\sum M_{c}= I \alpha_{BCD}$$

$<=> T*sen(53.13) * 5 = Iα + m*a(D)*sen(36,87)*5$

$<=> 4T = 416.67α + 150a(D)$

$$\sum F_{y}= (F(y))ef$$

$<=> T*sen(53.13) - P(BD) = m*a(Cy) - m*a(D)*sen(36.87)$

$<=> 0.8T - 490.5 = 50a(Cy) - 30a(D)$

$$\sum F_{x}= (F(x))ef$$

$<=> T*cos(53.13) = m*a(Cx) + m*a(D)*cos(36.87)$

$<=> 0.6T = 50a(Cx) + 40a(D)$

$a(C) = a(D)+α*r(CD) $

$= 0.8a(D)i-0.6a(D)j + αk * 5i$

$= 0.8a(D)i + (5α-0.6a(D))j$

and so i know that $a(Cx)=0.8a(D)$ and that $a(Cy)=5α-0.6a(D)$

and solving the system i got $T=-1032.66N$, $a(D)=7.745m/s^2$, $α=-7.125rad/s^2$, $a(C)=31.59m/s^2$, which are the wrong solutions. Comparing my equations between the FBD and KD with the resolution, my equations are wrong but i cant understand why

In the resolutions they use this equations but i still cant understand why

Answer 1

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

• equilibrium along the x-axis $$T\cos\theta = m \cdot a_{Cx} $$

• equilibrium along the y-axis $$T\sin\theta - m g= m \cdot a_{Cy} $$

• 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:

• $\vec{a}_D = \alpha_{\gamma, DE}\times \vec{r}_{ED}$
• $\vec{r}_{ED} = -3\vec{i} -4\vec{j} =\begin{bmatrix}-3\\-4\\0\end{bmatrix}$

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

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

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

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