1) a) The lady below is pulling on a 200 lb door that is on rollers and has a ce

Question

1) a) The lady below is pulling on a 200 lb door that is on rollers and has a center of gravity at the geometric center. If the door is slid opened at a constant acceleration and it takes 5 seconds to move it 12 feet, what is the acceleration?

b) If the lady is pulling on the door with a (constant) force of 50 lbs, determine the moment around the right roller MR in terms of the reaction force RL on the left roller.

c) What are the vertical reaction forces on each roller during the pull. (Hint: RL = RR = 100 lb if the door is not being pulled.)

d) What would the reaction forces be if the door handle was located at a height of six feet (along the line of the CG)?

Question 2

2) A bobble-doll sits on the dash of a small pleasure boat. The tiger head has mass = 100 grams and vibrates vertically in a spring with a constant   4 kN/m.

        a) what is the natural frequency of the bobble?

      b) In rough seas the boat is moving up and down on the waves according to the equation

                                d =   .7 sin (.2t)

            What is the equation for the vibration of the bobble head including the forced vibration of the boat?

c) Draw a Spectrum Diagram of the bobble-head vibration

Question 3

The carton above has a weight of 20 lbs and a coefficient of static friction of .3. If it is traveling on the conveyer at a (tangential) speed of 15 ft/sec, what is the smallest radius of curvature for the conveyor so the package does not slip.

Explanation / Answer

a)the acceleration is

a = (v^2/r)

where v = (12/5) ft/s and r = 6 ft

b)the vertical reaction force is

F = w x sinA

where w = 200 lb and A = tan^-1(12/6)

c)the reaction force is

F1 = w x sinA1

where A1 = tan^-1(6/6) = 45o

2)the angular frequency of the bobble is

w = (m/k)^1/2

where m = 100 x 10^-3 kg and k = 4 x 10^3 N/m

or f = 1/2pi x (m/k)^1/2

where f is frequency

equation for the vibration is

v = (ds/dt) = .7 x .2 x cos(.2t) = .14 x cos(.2t)

3)weight of carton w = 20 lbs

coefficient of static friction is u_s = .3

the speed is v = 15 ft/s = 15 x 0.3048 m/s

let r be the radius of curvature

we know that

u_s x g = (v^2/r)

or r = (v^2/u_s x g)

where g = 9.8 m/s^2

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