HINTS FOR PROBLEM SET #2

Problem Set #2 – Problem 1

Part a)

You want to find the acceleration, a, of a car entering a freeway from an on ramp.

Since you are told that the car accelerates uniformly you know that a is constant.  

Use the equation for uniformly accelerated rectilinear motion:

                 vfinal2 = vinitial2  +  2a (xfinal - xinitial)

Solve this for a.

Part b)

You want to get the time required for the car to reach 90 mph.

Use the equation for uniformly accelerated rectilinear motion:

                 vfinal = vinitial + at

Solve this for t.

Problem Set #2 – Problem 2

This is a problem where considering the relative motion will be helpful.

Box A and B have the same (constant) acceleration, a.

Thus the acceleration of box A with respect to the acceleration of box B is 0.

                 a_A/B = 0

It follows that the velocity of box A relative to box B, v_A/B, is constant. 

                 v_A/B = delta x_A/B / delta t = (30 - 18)ft / 1 sec  = 12 ft/sec.

(Note that the absolute velocities of A and B are not constant.

However, the relative velocity is constant.)

The absolute velocity of box A can be found by using:

                 v_A = v_B + v_A/B

At t = t_R,  v_B = 0  and  v_A/B = 12.

Plugging these into the equation gives:

                 v_A = 12ft/sec at t = t_R.

Now use the equation for uniformly accelerated rectilinear motion:

                 v_final = v_initial  + at

Considering block A at t = t_R this becomes:

                 12 = 0 + a t_R   

             or  

                 12 = a t_R                  (equation 1)

This is one equation in two unknowns, a and t_R.

You need another equation in these same two unknowns.

.

Use the equation for uniformly accelerated rectilinear motion:

                 x_final = x_initial  + (v_initial) t + 1/2 a t²

Considering block A at t = t_R this becomes:

                 18 = 0 + (0)t_R + 1/2 a t_R²     

             or

                 36  = a t_R²               (equation 2)

equation 2 can be used with equation 1 to solve for a and t_R

Problem Set #2 – Problem 3

This is a system with dependant motions.

There are two motions but only one degree of freedom.

You need to find the relationship between the positions and velocities of A, B, C and D.

Part a)

To want to get v_A.

Use the idea that the length of the entire pulley cord is constant.

You should be able to show that

               x_B + (x_B - x_A) + (-x_A) + (-x_A) = constant

or

              2x_B - 3x_A = constant
(If you need more explanation to see how this equation was obtained,

go to the bottom of this page for an explanation of 

"The Conservation of String".)

Differentiating this relationship between the positions
gives the relationship between the velocities:

               2v_B - 3v_A = 0

v_B is given so this can be solved for v_A.

Part b)

You want to get v_C.

Use the idea that the length of the pulley cord between points E and C is constant. 

This gives:

               x_B + x_B - x_C = constant

Differentiating gives:

               2v_B - v_C = 0

v_B is given so this can be solved for v_C.

Part c)

You want to get v_D.

Use the idea that the length of the pulley cord between points A and D is constant. 

This gives:

               -x_A - x_D = constant

Differentiating gives:

               -v_A - v_D = 0

v_A is known so this can be solved for v_D.

Part d)

You want to get the velocity of C relative to A.

This can be found from the relative velocity equation:
                v_C/A = v_C - v_A

Problem Set #2 – Problem 4

This is another system with dependant motions. 

There are three motions but only two degrees of freedom.

You need to find the relationship between the positions, velocities and accelerations of A, B and C.

Part a)

Using the idea that the length of the pulley cord is constant gives this equation:

              2x_A + x_B - 4x_C = constant

Differentiating this gives:

              2v_A + v_B - 4v_C = 0

              2a_A + a_B - 4a_C = 0

Solving this last equation for a_C gives:

              a_C = 0.5 a_A + 0.25 a_B     (equation 1)

a_A is given so you need to find a_B in order to solve this for a_C.

Since it is given that block B moves with constant acceleration,

you can use the equation for uniformly accelerated rectilinear motion:

              v_final² = v_initial²  +  2a (x_final - x_initial)

Plugging into this equation allows you to solve for a_B.

Now you can solve equation 1 for a_C (as a function of t).

Part b)

You want to find the distance C moves in 3 seconds.

You need to find x_C and v_C as functions of time, x_C(t) and v_C(t).

Since you have a_C as a function of t this is very straightforward. 

You just need to integrate using the initial conditions.

(i.e. at t = 0, v_C = 0 and x_C = 0)

But be careful!

The distance traveled from t=0 to t=3sec is not necessarily the position, x(t), at t=3sec. 
(Refer to Problem Set #1 – Problem 2.)

You need to determine if the velocity ever changed directions.

i.e. was there ever a time when v = 0? 

In this case the answer is yes and it occurs at t = 0.707 sec. 

Hence the distance traveled by C in 3 seconds is given by:

              |x(3) - x(0.707)| + |x(0.707) - x(0)|

"The Conservation of String"  Explained

What follows here is an explanation of how the relationships

between xA, xB, xC and xD in Problem 3 were obtained.
All use the idea that the cord (string) in inextensible,
that is, the length is constant.

For Part a of this problem you need to do is this:

1. Define the x position coordinate for each block, xA and xB.

    I chose the datum (origin) to be the wall on the left side
     and defined xA and xB to go from the wall to the points in the center of each pulley.
     (This makes the numbers easier.)

2. Divide the length of the cord into sections.

      first section is from point E to where it first touches the pulley on block B

      second section is from this point to other side (so it is a half circle)

      third section is from this point to the point where it first touches the pulley of block A

      fourth section is from this point to other side (another half circle)

      fifth section is from this point to the point where is first touches the pulley on the wall

      sixth section is from this point to the other side (another half circle)

      seventh section is from this point to the point on block A where the cord attaches

3. Now represent each section in terms of xA and xB.

      first section is just equal to xB

      second, fourth and sixths sections are all just constants
                 (= 1/2 pi x diameter of pulley), call them c2, c4 and c6

      third section is equal to (xB - xA)

      fifth section is equal to (c5 - xA)
                where c5 is a constant that represents the length from the wall at point E
                 to the center of the pulley mounted on the step

      seventh section is basically just like the fifth: (c7 - xA)

4. Add up all of these sections and set equal to the total length of the cord, Ctotal.

      xB + c2 + (xB - xA) + c4 + (c5 - xA) + c6 + (c7 - xA) = Ctotal

5. Put all of the constants on the same side.

6. You should get 2xB - 3xA = constant  

I noticed in the solutions for this set, the line right above this equation has an error.
There should be two (-xA) terms.

For parts b and c you need to define xC and an xD and go through the same procedure except only take the length of cord to the points C and D respectively.