Consider the movement in the vertical plane of an $AB$ bar with a mass of $20 kg$, $0.8 m$ long and supported by two cursors: one at the end A and the other at point G coinciding with the center of mass of the bar. Cursor A slides on the vertical guide at a constant speed $vA = 0.5 m / s$ (zero acceleration) due to a vertical force $F$ applied at point $A$. Cursor $G$ slides on a guide inclined at $45º$ to the horizontal.

For the instant when the bar is in the vertical position $(Θ = 0)$ calculate the force $ F $ and the reactions on the supports $A$ and $G$.

The solution is $F = 167.14 N; RA = 4.16N; RG = 23.5 N$

enter image description here

I already know that:

the angular velocity of the AB bar is $1.25 rad / s$

the acceleration of the center of mass is $0.88 m / s2$ and the angular acceleration of the $AB$ bar is $1.56 rad / s2$ (anticlockwise)

However i cant get the correct results for the reactions

$\sum M_{c}= I \alpha$ <=> $R_Gcos(45)*0.4=16/15*(1.56)$ <=>$R_G=5.88N$

$\sum F_{y}= (F(y))ef$ <=> $R_{Ax}+R_Gcos(45)=20*(-0.625)$

$\sum F_{x}= (F(x))ef$ <=> $R_Gsen(45)+F=20*(-0.625)$

then i got the wrong walues: $R_{Ax}=-16,657$, $R_{G}=5.88N$ and $F=-16.66N$

enter image description here

  • 1
    $\begingroup$ how can force F not cause an acceleration. Also with the bar at vertical position, it should turn counterclockwise. $\endgroup$ – kamran Mar 14 at 22:11
  • $\begingroup$ How did you arrive at the values for angular velocity and accelaration of bar AB? (I'm not suggesting that they are wrong, I merely want to see you line of thought). $\endgroup$ – NMech Mar 14 at 23:00
  • $\begingroup$ Which equations did you use? I am having trouble with the angular acceleration and the acceleration of the center of gravity. Did you use instantaneous centers? $\endgroup$ – NMech Mar 15 at 15:45
  • $\begingroup$ I don't quite understand how would there be reactions other than F = m*g, when the rod is in vertical position, as the points A and G change positions and slide freely along the path. $\endgroup$ – r13 Mar 15 at 19:26
  • 2
    $\begingroup$ I’m voting to close this question because the OP has vandalized their own question & closed their account. $\endgroup$ – Fred May 20 at 10:53

After verifying the values for angular velocity ($1.25 \left[\frac{rad}{s}\right]$)and acceleration ($-1.56 \left[\frac{rad}{s^2}\right]$), I calculated the

  • $\vec{v}_G$ the velocity of point G.

$$\vec{v}_G = 0.707\cdot \left(\cos(\frac{3\pi}{4}) \vec{i} + \sin(\frac{3\pi}{4}\vec{j})\right) $$

$$\vec{v}_G =-0.5\vec{i} + 0.5 \vec{j} =\begin{bmatrix} -0.5\\ 0.5\\0\end{bmatrix} $$

  • $\vec{a}_G$ the linear acceleration of point G:

$$\vec{a}_G = -0.884\cdot \left(\cos(\frac{3\pi}{4}) \vec{i} + \sin(\frac{3\pi}{4}\vec{j})\right) $$ $$\vec{a}_G = 0.625\vec{i} -0.625\vec{j} =\begin{bmatrix} 0.625\\ -0.625\\0\end{bmatrix} $$

enter image description here

So, point G is moving upwards, and the bar is rotating counter clockwise, at the point of interest, the linear and angular acceleration are "braking" point G and the rotation of the bar.

we take the equilibrium of moments about the center of the bar (G).

$$\sum M_{G}= I \alpha $$


  • $I = \frac{1}{12} m_{bar} L^2$


$$R_A *0.4=\frac{1}{12} m_{bar} L^2*(-1.56)$$

$$R_A =\frac{1}{12\cdot 0.4} m_{bar} (0.8)^2*(1.56)= -4.16[N]$$

Then x-axis:

$$\sum F_{x}= F_{x,ef} \Leftrightarrow $$

$$R_{G}\cos(45)+ R_{A}=20*(0.625)$$

$$R_{G} = \frac{1}{\cos(45)} \left(20*(0.625)- R_{A}) \right)$$ $$R_{G} = 23.56[N]$$

Finally we can proceed to the equilibrium of y axis:

$$\sum F_{y}= F_{y,ef} \Leftrightarrow $$

$$R_G \sin (45) + F -m_{bar}\cdot g= m_{bar}* a_{Gy}$$

$$ F = m_{bar}* (a_{Gy} + g) - R_G \sin (45) $$

$$ F = 20*(-0.625+9.81) - \frac{\sqrt{2}}{2} 23.56$$ $$ F = 167.04[N]$$

Final thoughts and advice

The biggest problem for deriving this is to realize what happens on the bar at the point of interest. Once you solve it analytically, it becomes clear. However the direction of the forces and acceleration is not (at least for me immediately) intuitive.

IMHO the best advice to you, is that you

  • make a clear free-body diagram and kinetic diagram, and always use the positive direction for the values you defined.

  • don't redraw the free body diagram with the "correct direction" (even if values become negative). The main problem is that if you do so, because you have cross products, the calculation is prone to errors.

Comparison to your solution

I could not make why:

  • you used $R_{Ax}$ in the equation for the equilibrium of force in the y axis.
  • you did not use the weight of the bar.
  • $\begingroup$ I suggest to have a closer look into DaveM2's response. I don't think the solution given by the author is correct (note the sum of forces is very close to the weight of the rod, F = mg = 20*9.81). $\endgroup$ – r13 Mar 17 at 22:19
  • $\begingroup$ I think, if it is solvable using the given conditions, the results must satisfy system equilibrium, both sum Fx and sum Fy are zero. If so, there must exist a force couple, M = Fgx x a, in which Fgx = -Fax, and a is the distance between blocks A and G, when the rod AB is in the vertical position. Note that the distance "a" is the key unknown, which requires more information be given. BTW, I could be wrong though. $\endgroup$ – r13 Mar 17 at 23:43

The problem seems irrational if the ends of the slide rails are anchored, and the G point is fixed at the midpoint of the AB bar. Point G must follow an arc but cannot because of the 45° slide bar. Even if it could, then at Θ=zero degrees the mechanism abruptly stops. Any kinetic energy at this point abruptly converts to a vertical force spike and the stopping time must be known to determine F. This is a shock problem if it is possible at all.


The sketch below is not a response to the question, but to show the final positions of the components (all in red), which may be helpful in deriving the correct solution, if any. Note that it is assumed that the overlap of support N and rod AB is feasible.

enter image description here


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