Find out the components of the force:
The frame of a certain machine accelerates to the right at 0.8 g m/sec2. As illustrated in Figure, it carries a uniform bent bar ABC weighing 50 N/m, pinned to it at C and braced by the consistent strut DE which weighs 100 N. Find out the components of the force at pin D.
Except for the application of inertia forces the solution of this problem is same as that for statics. The free body diagram of each of member is placed in dynamic equilibrium including inertia force acting through the C. G. of each of part, directed opposite to the acceleration. We should note that instead of locating C. G. for ABC, it is more suitable to apply the inertia reaction acting on each of part of its length. The inertia forces are
ma/g = 15 . a = (15/g) × 0.8 g = 12 N
ma = (25/g) . a = (25/g) × 0.8 g = 20 N
ma = (100/g) . a = (100/g) × 0.8 g = 80 N
Now Let us consider FBD of ABC.
Sum of moments about C shall eliminate three unknown forces and we get, directly Dx.
∴ ∑ M C = 0
- M AB × 15 - FAB × 50 - FBC × 25 + Dx × 35 = 0
(MBC has not been taken because this is acting vertically which passes through point C).
∴ - 15 × 15 - 12 ×50 - 20 × 25 + Dx × 35 = 0
∴ Dx = 37.8 N
Referring to FBD of DE now in which Dx is known, we take sum of moments about E which shall remove Ex and Ey.
∴ ∑ M E = 0
∴ - 37.8 × 35 - 80 × 17.5 - 100 × 12.5 + Dy × 25 = 0
∴ Dy = 158.92 N
If needed, components of reactions at C and E may be determined by applying
∑ Fx = 0, ∑ Fy = 0 , to each body separately.
The body of mass 100 kg is supported by wheels at B which freely roll without friction and by a skid at A under which the coefficient of friction is equal to 0.5. Calculate the value of P to cause an acceleration of (g/4) m / sec2 .