Mechanics

Contents

Applications

# Statics

### Lid of a roof scuttle

Steps to follows:
- Understand the mechanism,
- Draw all the forces
- Draw the FBD,
- Use ∑ F = 0 and ∑ M(F)any-point = 0.

We have three forces:
A) The force exerted by the rod CD: FCD.
B) The reactions at the hinges A and B:
- Recation on A: RA, and
- Reaction on B: RB, and

Since the hinge at B does not exert any axial thrust, RB,x (component over x) is zero.

We have:
RA = RA,x i + RA,y j + RA,z k
And
RB = RB,y i + RB,z j

∑ F = 0
RA + RB + FCD + P = 0

Over x:
∑ Fx = 0
= RA,x + RB,x + FCD,x + Px = 0
0 + 0 + 0 + 0 = 0

Over y:
∑ Fy = 0
= RA,y + RB,y + FCD,y + Py
= RA,y + RB,y + FCD,y - P

Over z:
∑ Fz = 0
= RA,z + RB,z + FCD,z + Pz
= RA,z + RB,z + FCD,z + 0

∑ M[F]/A = 0
M[RA]/A + M[RB]/A + M[FCD]/A + M[P]/A = 0

M[RA]/A = 0

M[RB]/A = (b,0,0) x(0, RB,y, RB,z) =
(0,- b RB,z, b RB,y)

M[FCD]/A = (0,AD sin α,AD cos α) x (0, FCD,y , FCD,z) =

M[P]/A = AG x P = (b/2,(a/2)sin α ,(a/2)cos α ) x (0, - P, 0) =
(1/2) (P a cos α ,0, - P b)
∑ M[F]/A = 0
= 0 +
(0, - b RB,z, b RB,y) +
(AD sin α FCD,z - AD cos α FCD,y, 0,0 ) +
(1/2) (P a cos α , 0, - P b)

Then:
AD sin α FCD,z - AD cos α FCD,y + (1/2) P a cos α =
0 - b RB,z = 0 → RB,z = 0

RB,z = 0

b RB,y - (1/2) P b = 0 → RB,y = P/2

RB,y = P/2

Over y:
RA,y + FCD,y = P/2

Over z:
RA,z + FCD,z = 0

AD sin α FCD,z - AD cos α FCD,y + (1/2) P a cos α = 0

FCD,z = - FCD cos β
FCD,y = + FCD sin β

AD sin α (- FCD cos β) - AD cos α (+ FCD sin β) + (1/2) P a cos α = 0

AD FCD [- sin α cos β - cos α sin β] + (1/2) P a cos α = 0

AD FCD [sin α cos β + cos α sin β] = (1/2) P a cos α

AD FCD sin (α + β) = (1/2) P a cos α
Thus:
FCD = P a cos α /2 AD sin (α + β)

FCD = P a cos α /2 AD sin (α + β)

And:
RA,z = - FCD,z = FCD cos β
= P a cos α cos β /2 AD sin (α + β)

RA,z = P a cos α cos β /2 AD sin (α + β)

and:

FCD,y = + FCD sin β =
P a cos α sin β /2 AD sin (α + β)

RA,y + FCD,y = P/2
RA,y = P/2 - FCD,y =
P/2 - P a cos α sin β /2 AD sin (α + β)

RA,y = P/2 - P a cos α sin β /2 AD sin (α + β)

To find AD and the angle β, we can use the sine law:
d/sinα = c/sin(α + &β) = AD/sin β
Then:
sin(α + β) = c sinα/d α + β = arcsin [c sinα/d]
Thus:
β = arcsin [c sinα/d] - α

β = arcsin [c sinα/d] - α
And
AD = d sinβ /sin α = d sin{arcsin [c sinα/d] - α} /sin α

AD = d sinβ /sin α =
d sin{arcsin [c sinα/d] - α} /sin α

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