Question 11.13: Leaning panel A rectangular panel ABCD of mass M is (rather ...
Leaning panel
A rectangular panel ABCD of mass M is (rather carelessly) placed with its edge AB on the rough horizontal floor z = 0 and with the vertex D resting against the smooth wall x = 0, as shown in Figure 11.13. The four vertices of the panel are at the points A(2, 0, 0), B(6, 4, 0), C(4, 6, 6) and D(0, 2, 6) respectively. Given that the panel does not slip on the floor, find the reaction force exerted by the wall.
The external forces acting on the panel are the normal reaction of the wall Pi, the weight force −Mgk, and the reaction of the floor on the edge AB. Now the reaction of the floor is distributed along the edge AB and, although we could treat it as such, this is an irrelevant complication. We will therefore suppose that (in order to avoid damage to the floor) the panel has been supported on two small pads beneath the corners A and B, in which case the reaction of the floor consists of the forces X and Y as shown.
We now apply the equilibrium conditions. The condition F = 0 yields
N i + X + Y -M g k = 0. (11.29)
If we take moments about the corner A, the reaction X makes no contribution and the condition K _{A}= 0 becomes
\overrightarrow{A D} \times(N i )+\overrightarrow{A B} \times Y +\overrightarrow{A G} \times(-M g k )= 0.
On inserting the given numbers, this equation becomes
N(6 j -2 k )+(4 i +4 j ) \times Y -M g(3 i – j )= 0. (11.30)
The six scalar equations in (11.29), (11.30) contain the seven scalar unknowns X, Y, N which means that the problem is not statically determinate. However, this does not stop us from finding N, since Y can be eliminated from equation (11.30) by taking the scalar product with the vector 4i+4 j. (This is equivalent to taking moments about the axis AB so that both X and Y disappear to leave N as the only unknown.) This gives the reaction exerted by the wall to be N = Mg/3.
