Mechanics - Inclined plane, and projectles

In summary: Pcosθ - FB - T - 2mgsin30 = (√3)/6*mgT - FA - mg sin30 = (√3)/3*2mgSolving for T, we get:T = (√3)/6*mg + Pcosθ - FB - 2mgsin30T = (√3)/6*mg + (√3)/3*sinθ*(P + 2mg)T = (√3)/6*mg + (√3)/3*sinθ*(P + 2mg)Substituting this into the second equation, we get:T
  • #1
Thomas154321
3
0
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 [Broken]
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.
 
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  • #2


1. For part a), we can use the equations you have already derived to solve for T. We know that the particles will move upwards along the line of greatest slope if the net force in that direction is positive. Therefore, we can set the equation from step 1 equal to 0 and solve for T:

Pcosθ - FB - T - 2mgsin30 = 2ma
0 = Pcosθ - (√3)/3*(Psinθ + 2mg cos30) - T - 2mgsin30
0 = P(cosθ - (√3)/3*sinθ) - 2mg((√3)/3*cos30 + sin30)
0 = P(cosθ - (√3)/3*sinθ) - 2mg(√3)/6
0 = P(cosθ - (√3)/3*sinθ) - (√3)/3*mg

Now, we can use the equation from step 3 to solve for T:

T = ma + 3mg/4
T = 0 + 3mg/4
T = (3√3)/4*mg

Substituting this into the first equation, we get:

0 = P(cosθ - (√3)/3*sinθ) - (√3)/3*mg
P = (√3)/3*mg/(cosθ - (√3)/3*sinθ)
P = (√3)/3*mg/(√3/2 - (√3)/3*1/2)
P = (2√3)/3*mg/(√3/2)
P = (4√3)/3*mg

Therefore, the least magnitude of P for which the two particles move upwards along the line of greatest slope is (4√3)/3*mg, and it acts in the direction of the line of greatest slope.

2. For part b), we can use the same approach, but instead of setting the net force equal to 0, we want to set it equal to the maximum possible force that will not cause the particles to slip downwards. This force is determined by the maximum possible friction force, which is equal to the coefficient of friction multiplied by the normal force. In this case, the normal force is equal to the weight of the particle (mg for A and 2mg for B).

Therefore, we can
 
  • #3


2: First, we can find the equations of motion for the missiles using the equations: s=ut + 1/2at^2 and v=u+at, where s is the displacement, u is the initial velocity, a is the acceleration, and t is time.

For missile A:
Vertical motion: sA=100(sin arcsin(3/5))t - 1/2(10)(t^2)
Horizontal motion: xA=100(cos arcsin(3/5))t

For missile B:
Vertical motion: sB=200(sin arcsin(4/5))(t-T) - 1/2(10)((t-T)^2)
Horizontal motion: xB=200(cos arcsin(4/5))(t-T)

Next, we can set the equations for the vertical motion of the missiles equal to each other, since they will collide at the same height:
100(sin arcsin(3/5))t - 1/2(10)(t^2) = 200(sin arcsin(4/5))(t-T) - 1/2(10)((t-T)^2)

Simplifying and rearranging, we get:
5T^2 + (10t-160)T -100t = 0

This is the same equation you got, but we can now solve for T using the quadratic formula:

T = (-b ± √(b^2 - 4ac))/2a

Plugging in the values, we get:
T = (-10t + 160 ± √((10t-160)^2 - 4(5)(-100t)))/(2(5))

Simplifying, we get:
T = (-10t + 160 ± √(100t^2 - 3200t + 25600 + 2000t)) / 10
= (-10t + 160 ± √(100t^2 - 1200t + 25600)) / 10
= (-10t + 160 ± (10√(t^2 - 12t + 256))) / 10
= (-t + 16 ± √(t^2 - 12t + 256)) / 10

Since T represents time, it cannot be negative, so we can take the positive solution:
T = (-t + 16 + √(
 

1. What is an inclined plane and how does it work?

An inclined plane is a simple machine that consists of a flat surface with one end higher than the other. It works by reducing the amount of force needed to lift an object by spreading it out over a longer distance. This allows for heavier objects to be moved with less effort.

2. How does the angle of the inclined plane affect its mechanical advantage?

The mechanical advantage of an inclined plane is directly related to the angle of the plane. The smaller the angle, the longer the distance over which the force is applied, resulting in a greater mechanical advantage. However, a steeper angle also requires a longer inclined plane to achieve the same mechanical advantage as a shallower angle.

3. What is the formula for calculating the mechanical advantage of an inclined plane?

The mechanical advantage of an inclined plane can be calculated using the formula MA = Length of plane/Height of plane. This means that the greater the length of the plane compared to its height, the greater the mechanical advantage.

4. How does the weight of an object affect its movement on an inclined plane?

The weight of an object has a direct impact on its movement on an inclined plane. The greater the weight of the object, the more force is required to move it up the plane. However, once the object is on the inclined plane, the angle and length of the plane will determine the amount of force needed to maintain its position.

5. How is the motion of a projectile affected by gravity?

Gravity has a significant effect on the motion of a projectile. It causes the projectile to follow a curved path, known as a parabola, due to the constant downward acceleration. This acceleration also affects the speed of the projectile, causing it to increase as it moves downward and decrease as it moves upward.

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