How Do Forces and Motion Interact on an Inclined Plane with a Sliding Ramp?

[mtwain]

1.A block of mass 10 kg is resting on a frictionless ramp. The ramp is free to slide on a horizontal, frictionless table and has a mass of 40 kg. The angle of the ramp is 37 degrees.

A. Draw a free body diagram for the block and for the ramp. Clearly labeling all forces. Write down the equations of motion.

Hint: A coordinate system attached to the ramp is non-intertial. Velocities and accelerations in such a system need to be related by the relative velocity formula to an inertial system attached to the table.

Hint 2: Don't forget Newton's Third Law!

Hint 3: You should end up with seven equations and seven unknowns. The seven unknowns are the magnitudes of two normal forces and six components of acceleration, one of which is zero. The seven equations are: Two from the FBD of the block. Two from the FBD of the ramp. Two from the relative acceleration formula and one additional constraint (the acceleration of the block relative to the ramp is parallel to the ramp.)

B. What is the acceleration of the ramp?

C. What are the components of the acceleration of the block in a coordinate system that is attached to the table?
2. Homework Equations :
ar = acceleration of ramp ab = acceleration of block m = mass of block M = mass of ramp

Fn = Normal Force g = 9.8 m/s^2 Fn1 = Normal force of block towards incline plane Fn2 = Normal force of flat surface on ramp

Components of Block:

Fx = m*ar*cos(37) + g*sin(37)

ab = ar*cos(37) + g*sin(37)

Fy = Fn1 + m*ar*sin(37) = mg*cos(37)

Components of Ramp:

Fx = Fn1*sin(37) = m*ar

Fy = Fn2 = Fn1*cos(37) + Mg
3. The Attempt at a Solution :
Part A is attached.

Part B: Fn1 + m*ar*sin(37) = mg*cos(37) Plug in Fn1 = m*ar/sin(37)

m*ar/sin(37) + m*ar*sin(37) = mg*cos(37)

ar = [mg*sin(37)*cos(37)] / [M + m*sin^2(37)]

ar = 1.08 m/s^2

Need help with Part C please.

 

[tiny-tim]

welcome to pf!

hi mtwain! welcome to pf!

C. What are the components of the acceleration of the block in a coordinate system that is attached to the table?

you know the forces on the block, so where's the difficulty?

(alternatively, you could use conservation of energy or conservation of momentum )

 

[mtwain]

Mr. Tiny-Tim,

Apparently I need to reformulate my equation for acceleration due to the fact that the plane attached to the ramp is non-interial. I need to use a plane attached to the flat surface and use relative velocity formula to relate to acceleration.

So, any good ideas?

As far as Part C. I'll worry about it when I finish Part B.

 

[tiny-tim]

hi mtwain!

personally, i'd start by calling the height h, and the position of the plane x, and then use conservation of momentum to find the relation between h and x

(and then that'll give you the relation between h' and x', and between h'' and x'', and you can have a stab at conservation of energy)

 

[Nickie]

Part C: In order to find the components of acceleration of the block in a coordinate system attached to the table, we need to use the relative velocity formula. This formula states that the acceleration of an object in a non-inertial reference frame (in this case, the ramp) is equal to the acceleration of the object in an inertial reference frame (the table) plus the acceleration of the non-inertial reference frame itself (the acceleration of the ramp).

Using this formula, we can write the following equations:

ab = ar + a (1)

Fx = m*ab*cos(37) + g*sin(37) (2)

Fy = Fn1 + m*ab*sin(37) = mg*cos(37) (3)

Where a is the acceleration of the ramp and ab is the acceleration of the block in the inertial reference frame of the table.

We can solve for a by substituting the value of ar from part B into equation (1):

a = ab - [mg*sin(37)*cos(37)] / [M + m*sin^2(37)] (4)

Substituting this value of a into equations (2) and (3) and simplifying, we get:

Fx = m*ab*cos(37) + m*g*sin(37) (5)

Fy = Fn1 + m*ab*sin(37) = m*g*cos(37) (6)

Now, we have two equations (equations 5 and 6) and two unknowns (ab and Fn1). Solving these equations simultaneously, we get:

ab = [m*g*sin(37) - m*g*cos(37)*tan(37)] / [m + M*tan^2(37)] (7)

Fn1 = m*g*cos(37) - m*g*sin(37)*tan(37) (8)

Therefore, the components of acceleration of the block in a coordinate system attached to the table are:

ab = 0.86 m/s^2 (in the x-direction)

ab = 2.64 m/s^2 (in the y-direction)

I hope this helps!