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I am solving problems from the book: Meriam Kraige Engineering Mechanics statics vol 1. The problem is as follows:

The sliding door rolls on the two small lower wheels $A$ and $B$. Under normal conditions, the upper wheels don't touch their horizontal guide. The question is to find out the force $P$ required to slide the door at a steady speed if wheel $A$ becomes "frozen" and doesn't turn in its bearing. The coefficient of kinetic friction between a frozen wheel and the supporting surface is $0.3$ and the center of mass of the 64 kg door is at its geometric center. We have to neglect the small diameter of the wheels.

enter image description here

My attempt

I calculated the normal reactions at the two wheels to be equal to $32g$ each. I then applied $P=0.3N$ which gave me the incorrect value of $P$ .I couldn't get where I am wrong.Any help shall be highly appreciated.Thanks.

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  • $\begingroup$ I believe the reactions are not the same. $\endgroup$
    – Jem Eripol
    Sep 23 '17 at 7:05
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See picture below for reference. The thing is, the forces are now applied at a body. This is quite different when you are analyzing a particle because the position of each force are significant enough to change the effects of other forces.

Reference drawing.

Notes:

  1. The inertia is neglected because the velocity is to be constant (steady speed)
  2. The friction force, $F_f$, depends linearly on the normal force, $R_a$. the relationship is given by: $F_f=\mu\ N$
  3. Since the point of contact is at point $A$, the friction force acts at point $A$ as well.
  4. The friction in the rotating wheel, $B$ is neglected.

So to sum up, you need to consider the reduction of reaction force at A due to turning moment of force $P$. Such that:

Summing up forces at $B$, gives:

$$P\ (1)\ -\ 64\ (0.35)\ +\ R_a\ (0.7)\ =\ 0$$

and summing up forces along x-axis gives:

$$P\ =\ F_f\ =\ \mu_k\ R_a$$

$$P\ = 0.3\ (R_a)$$

Solving the two equations simultaneously gives:

$$P\ = \ 6.72\ kg$$

$$R_a\ = \ 22.4\ kg$$

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  • $\begingroup$ is it the same with the answer on the book? $\endgroup$
    – Jem Eripol
    Sep 23 '17 at 7:33
  • $\begingroup$ Yes it is the same $\endgroup$
    – user471651
    Sep 23 '17 at 9:51

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