Two Mechanics Questions: Projectile and Inclined Plane

[Thomas154321]

1) Two particles, A and B, of masses m and 2m, respectively,are placed on a line of greatest slope, l, of a rough inclined plane which makes an angle of 30 degrees with the horizontal. The coefficient of friction between A and the plane is (√3)/6 and the coefficient of frictuon between B and the plane is (√3)/3. The particles are at rest with B higher up l than A and are connected by a light inextensible string which is taut. A force P is applied to B.

a) Show that the least magnitude of P for which the two particles move upwards along l is (11√3)mg/8 and give, in this case, the direction in which P acts.

B) Find the least magnitude of P for which the particles do not slip downwards along l.

http://img243.imageshack.us/img243/8022/mechanics1uy8.png
I think it's all in the diagram except the angle of inclination of 30. The angle between P and the plane is θ.


1: Resolving parallel to the slope at B gives: Pcosθ - FB - T - 2mgsin30 = 2ma.

2: Resolving perpendicular to the plane at B gives: RB = Psinθ + 2mg cos30.

3: Resolving parallel to the slope at A gives: T - FA - mg sin30 = ma.

4: Resolving perpendicular to the plane at A gives: RA = mg cos30.

FB = (√3)/3*RB
= (√3)/3*(Psinθ + 2mg cos30).

FA = (√3)/6*RA
= mg/4.

From eq. 3: T = ma + FA + mg/2
= ma + 3mg/4.



Now I'm a bit confused - I have 8 unknowns - a, P, θ, FA, FB, RA, RB, T - and only 6 equations. For part b) I can put a=0 and F=uR, but in part a), a =/ 0. I don't know what to do next. Any help please?











2) The points A and B are 180 metres apart and lie on horizontal ground. A missile is launched from A at speed 100m/s and at an acute angle of elevation to the line AB of arcsin (3/5). A time T seconds later, an anti-missile missile is launched from B, at speed 200m/s and at an acute angle of elevation to the line BA of arcsin (4/5). The motion of both missiles takes place in the vertical plane containing A and B, and the missiles collide.

Taking g=10m/s and ignoring air resistance, find T.



I'm really not too sure on how to solve this, so I've just calculated the heights of each and said that these must be equal.

Vertically at A: a=-g, u=100*(3/5), s=yA
Using s=ut + 1/2at^2: yA = 100*(3/5)t -5t^2.

Vertically at B: a=-g, u=200*(4/5), s=yB
Using s=ut + 1/2at^2: yB = 200*(4/5)(t-T) - 5(t-T)^2


100*(3/5)t -5t^2 = 200*(4/5)(t+T) - 5(t+T)^2
=> 60t - 5t^2 = 160(t+T) - 5(t+T)^2
=> 60t - 5t^2 = 160t + 160T - 5t^2 -10tT -5T^2
=> 5T^2 + (10t-160)T -100t = 0.


This equation doesn't seem to get me anywhere. Can someone explain what I should be trying to do? Thanks.

[VBAhack]

Re the projectile problem, I think you are on the right track. A couple of comments.

1. I think you inadvertently changed (t-T) to (t+T):

Using s=ut + 1/2at^2: yB = 200*(4/5)(t-T) - 5(t-T)^2
100*(3/5)t -5t^2 = 200*(4/5)(t+T) - 5(t+T)^2

2. What you have is 1 equation with 2 unknowns (t and T). I think you need a 2nd equation. Isn't it also true that the sum of the horizontal distances traveled must equal the original separation of 180m? Since both missles have a parabolic trajectory, there are potentially 2 times the heights are the same. Drawing a picture always helps.

P.S. did you ever solve the other problem?
http://www.vbforums.com/showthread.php?t=429047

[Thomas154321]

Ah thanks. I'll try that!

And yes, I did the other question all by my self and got a perfect solution! My teacher thought it was "very impressive work". Only problem was in 5 pages of work, I forgot to simplify the mass ratio from 6/10 to 3/5. Anyway, your help with that one was also helpful, so thanks again. :)

[VBAhack]

I managed to get 1a (show that P = (11√3)mg/8). Here is the methodology I used:

1) consider the 2 masses connected by a string as a single unit (i.e. don't bother to break the string apart in the FBD)

2) Just barely moving upwards means that the component of P along the incline pulling the objects up is equal to the sum of forces resisting the masses moving up (gravity and friction)

3) Assume that P acts at some angle to the incline - which you've already done. The means there is a component of P acting on B along the incline (pulling B and A up) and a component of P acting on B in the normal direction of the incline (affecting frictional resistance of B).

4) By summing forces along the incline, obtain an expression for P in terms of θ (i.e. the angle that P makes with the incline) and mg.

5) Differentiate P with respect to θ, set equal to zero, and solve for θ (if you want to minimize something, chances are a derivative is involved!)

6) With θ from 5) calculate the value of P.

It's a bit tedious, but works. ;)

P.S. I don't know where these problems come from, but they are excellent!

[Thomas154321]

Here's what I think is the solution for Question 1:

http://img268.imageshack.us/img268/520/readimagephptj2.gif

[zaza]

A comment:

If P acts at an angle compared to the slope, then there must be a component to P which acts perpendicular to the slope. This component of the force is therefore not contributing anything to the movement of the masses. Hence the magnitude of the force must be bigger than it needs to be, since this component of it is unused.

You are assuming that the string can contribute something to the frictional force on one of the masses, ie that it is lifting it off the surface. I do not think that this is the intention of the question; it states "...a light inextensible string..." which I think is meant just to mean a motive force.


You don't need to differentiate anything, the minimum value of P is the point at which the forces balance.


IMHO.


zaza

[VBAhack]

Here's what I think is the solution for Question 1:

Looks good - same result for both as I got. I like the way you avoided taking a derivative by simplifying cosθ + sinθ/√3. and cosθ - sinθ /√3. My method wasn't as elegant:

(1) P = (11mg)/4(cosθ + sinθ/√3)

(2) To minimize P, maximize: cosθ + sinθ/√3

(3) derivative = 0: -sinθ + cosθ/√3 = 0

(4) sinθ = cosθ/√3

(5) tanθ = 1/√3 ==> cosθ = √3/2 ==> sinθ = 1/2

(6) cosθ + sinθ/√3 = √3/2 + 1/2√3 = 2/√3

(7) P = (11√3)mg/8

These are fun! Have any more? :D

To zaza: I thought the same thing as you about P acting only along the incline but the result wasn't as stated in the problem. If you assume P acts at an angle to the incline (this is implied in the problem statement), lo and behold there is an angle that minimizes the value of P, since it acts directly on B and contributes to the frictional resistance (only for B). That's what makes this an interesting problem. :) The string doesn't contribute - it only connects A and B.

[zaza]

The string doesn't contribute - it only connects A and B.

Whoops - misread it. Thought there was a string at P as well...

[Thomas154321]

I'll probably get some more on Friday - my teacher only gives me one or two a week.