The rigid bar AD is supported by two steel wires of 1 16-in. diameter (E 5 29 3 106 psi) and a pin and bracket at A. Knowing that the wires were initially taut, determine (a) the additional tension in each wire when a 220-lb load P is applied at D, (b) the corresponding deflection of point D.

Rhe image is missing and i have attached it

A) Tension in A (Pa) = 121.38N and Tension in B (Pb) = 75.86N

B) Deflection at B = 13.64 x 10^(-3) inches

Explanation:

A) Diameter = 1/16 inches.

Area = π(1/16)²/4 = π/1024

Deflection is given by a formula ;

δi = PLi/EA

A = π/1024

E = 29 x 10^(6) psi

Li is length AE on the image = 15 inches.

Thus,

δA = (Pa x 15)/(29 x 10^(6) x (π/1024))

Similarly,

δC = (Pc x 8)/(29 x 10^(6) x (π/1024))

δA = (24/16)δB = (3/2)δB

δC = 8/16δB = (1/2)δB

Thus,

(3/2)δB = (Pa x 15)/(29 x 10^(6) x (π/1024))

δB = 2/3[(Pa x 15)/(29 x 10^(6) x (π/1024))]

Also,

(1/2)δB = (Pc x 8)/(29 x 10^(6) x (π/1024))

δB = 2[(Pc x 8)/(29 x 10^(6) x (π/1024))]

Equating both δB, we have,

2/3[(Pa x 15)/(29 x 10^(6) x (π/1024))] = 2[(Pc x 8)/(29 x 10^(6) x (π/1024))]

Thus,

(2/3)Pa x 15 = 2(8Pc)

So,

Pa = (8/5)Pc

Thus,

(8/5)Pc(3/2) + (1/2)Pc = 220

Multiply through by 10;

24Pc + 5Pc = 2200

29Pc = 2200

Pc = 2200/29 = 75.86N

Pa = (8/5)Pc

So, Pa = (8/5) x 75.86 = 121.38N

B) from earlier,

δC = (Pc x 8)/(29 x 10^(6) x (π/1024))

So, δC = (75.86 x 8)/(29 x 10^(6) x (π/1024)) = 6.82 x 10^(-3)

Also,we saw that δC = (1/2)δB

Thus,

6.82 x 10^(-3) = (1/2)δB

δB = 2 x 6.82 x 10^(-3) = 13.64 x 10^(-3) inches

f=ma newton

f=100x2.5=250n

when two balls of the different masses are dropped from equal height, they reach the ground at the same time .this is because they experience the same acceleration.

the acceleration is independent of mass.

thus when they are dropped they will always maintain the same velocity and travel the same distance (neglect air resistance of course) at the same time.