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enter image description here

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:

enter image description here

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

enter image description here

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  • $\begingroup$ I thought about starting to edit your question but it was really painful. So I opted (eventually) for just doing the answer. Again (this is not the first time), please make an effort to use a more commonly accepted format for your questions. It will make it a lot easier for other people to read it and reply to it. $\endgroup$
    – NMech
    Mar 3 at 10:23
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    $\begingroup$ Please note, deleting a question because you didn't get the answers that you liked within your own time frame is not acceptable here. Questions and answers stay open for ever to help future users. Do not edit your question to delete it. $\endgroup$
    – hazzey
    May 20 at 13:37
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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

  • $a_B = 1.61260\mathbf{\vec{i}} - 14.110\mathbf{\vec{j}}$
  • $a_C = 1.61260\mathbf{\vec{i}} - 7.659863$
  • $a_D = 1.61260\mathbf{\vec{i}} - 1.2094520\mathbf{\vec{j}}$
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  • $\begingroup$ Because when you take the moment with respect to C, you consider forces acting on the bar BCD. So effective force is the acceleration of the bar BCD and its considered on point C. $\endgroup$
    – NMech
    Mar 3 at 11:31
  • $\begingroup$ When you are writing the equations of equilibrium you are considering the equilibrium of the bar BCD. Therefore you write the equations about the center of gravity C. The acceleration of point B and point D are directly related to the acceleration of point C (which you already considered) and the angular accleration. $\endgroup$
    – NMech
    Mar 3 at 11:50

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