I need to solve for the mass of block C that is in the verge of slipping. All surfaces have a coefficient of static friction of 0.11. The cable connecting the wedge B and C is parallel with the incline. I have given for the mass of the wedges A and B. enter image description here

I have established the FBD and its initial equations but I am not sure if it is correct.

For wedge A,

enter image description here

The initial equations are $\sum F_x = F_1(cos20) + N_1(sin20) - F_2(cos40) - N_2(sin40) = 0$ and $\sum F_y = W_A - F_1(sin20) + N_1(cos20) - F_2(sin40) + N_2(cos40) = 0$

For wedge B,

enter image description here

The initial equations are $\sum F_x = F_3 + F_2(cos40) + N_2(sin40) + T(cos11) = 0$ and $\sum F_y = -W_B - T(sin11) + F_2(sin40) - N_2(cos40) = 0$

For block C,

enter image description here

My initial equations are $\sum F_x = -T - W_C(sin11) - F_4 = 0$ and $\sum F_y = N_4 - W_C(cos11) = 0$

I will edit my illustration for my tension here. I think it is wrong.

Did I miss something in the cases of friction? Are those all I need to solve for the mass of block C?

  • $\begingroup$ Is the first figure drawn by you ? If so, can you post the original figure also in the question? $\endgroup$
    – AJN
    Jun 10 at 16:17
  • $\begingroup$ The direction of We of block C. Can you plot that vector on the first diagram? Will it look like the one drawn in the last diagram? Same thing with the tension when drawn on the last diagram. Plot both in the first figure and compare to the last one. String is parallel to incline and so it is parallel to the base of block C. But it doesn't look like that in the last diagram??? $\endgroup$
    – AJN
    Jun 10 at 16:21
  • $\begingroup$ Angle between Wb and T in B's FBD is 90-11 deg. I expected it to be 90+11 deg in the FBD for C block. But it looks like 90 deg. (I assume that Weight are all parallel). $\endgroup$
    – AJN
    Jun 10 at 16:26

Your work (free body diagrams and equilibrium equations) contents mistakes. The sketch below shows the force diagram of block "C" as an example:

enter image description here

$\sum Fy = 0$, the normal force, $N_4 = W_Ccos\theta$

The shear friction force, $S_4 = \mu N$

$\sum Fx = 0$, the required tension, $T = W_Csin\theta - S_4$

You shall get the tension produced by the blocks "A" and "B", in the same manner as shown above, for which the magnitude shall be the same as $"T"$.

Note that the forces $T, W,$ and $N$ must act thru the centroid of mass "C". The shear friction force $S$ is a reactive force, its magnitude depends on the normal force and friction coefficient only, and always in the direction against the motion.


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