[20分] 2條物理AL靜態平衡

1. Let Fa and Fb be the forces at A and B respectively. Since the floor is smooth, Fa and Fb are vertically upward.
hence for equilibrium, upward force = downward force
854 = Fa + Fb -------------- (1)
Let the mid-point of BD be M and the mid-point of AB be M', and the tension of the tie rod be T. Consider the portion of the ladder CE, taking moment about C,
T x MC = Fb x M'C -------------------- (2)
where MC and M'C can be found from simple geometry
Now, consider the portion AC, let the point where the man stands be N. Taking moment about C again,
T x MC + 854 x (perpendicular distance of man's weight from C) = Fa x AM' ---------------------- (3)
Since the [perpendicular distance of man's weight from C] can also be found from simple geometry, you can solve the three equations for the three unknowns Fa, Fb and T
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2. The crucial point in this problem is the centre of gravity of the block(s) above a particular block must not exceed the edge of that block.
Consider the upper most two blocks (block 1 and block 2, say), the centre of mass of the upper most block (block 1) must be at most at the edge of the second upper most block (block 2). Hence, a1 = L/2 (as the centre of mass of the upper most block is at L/2 from its edge).
Let m be the mass of each block and x be the centre of mass of [block 1 + block 2] from the edge of block 2, thus
m.x = m.(L/2- x)
solve for x gives x = L/4
i.e. the centre of mass of [block 1 + block 2] combined togerther is at a distance of L/4 from the edge of block 2.
When [block 1 + block 2] is placed on top of block 3, the centre of mass of [block 1 + block 2] must not exceed the edge of block 3. Thus a2= L/4
Let the combined centre of gravity of [block 1 + block 2+block 3] be x' from the edge of block 3, thus
2m.x' = m.(L/2-x')
solve for x' gives x' = L/6

2009-07-12 00:07:18 補充：
When these 3 blocks are placed on top of block 4, the combined centre of gravity of the 3 blocks must not exceed the edge of block 4. That is, a3 = L/6

2009-07-12 00:07:49 補充：
Likewise, le x'' be the combined centre of gravity of [block 1+block 2+block 3+block 4] from the edge of block 4, then
3m.x'' = m.(L/2-x'')
solve for x'' gives x'' = L/8

2009-07-12 00:08:22 補充：
Since the centre of gravity of these four blocks must not exceed the edge of the table, i.e. a4 = L/8
Therefore, we have,
a1 = L/2 ; a2 = L/4 ; a3 = L/6 ; a4 = L/8
h = a1 + a2 + a3 + a4 = L[1/2+1/4+1/6+1/8] = 25L/24