# Force P required to slide the door if one of the wheel becomes frozen

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.

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.

• I believe the reactions are not the same. Sep 23 '17 at 7:05

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.

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:

$$P\ = \ 6.72\ kg$$

$$R_a\ = \ 22.4\ kg$$

• is it the same with the answer on the book? Sep 23 '17 at 7:33
• Yes it is the same Sep 23 '17 at 9:51