How to Find the Acceleration of a Block and an Inclined Plane?

In summary, a hard inclined plane problem is a physics problem that involves calculating the forces and motion of an object on an inclined plane, taking into account factors such as friction, gravity, and the angle of the incline. To solve it, you need to identify and draw all the forces, and use trigonometry and Newton's laws of motion. The main difference between a hard and easy inclined plane problem is the complexity of factors involved. The coefficient of friction can be determined experimentally, and some real-world applications of hard inclined plane problems include analyzing the motion of objects on ramps and calculating the angle of wheelchair ramps.
  • #1
bon
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Homework Statement



so a block of mass m slides on a frictionless surface of an inclined plane of angle theta which itself has mass M and can slide on a horizontal surface. Assuming no friction, find the accel of the block and the inclined plane..





The Attempt at a Solution



well i see that there are two components to the accel of the block and one component to the accel of plane, giving 3 equations...but how do i solve to find the accel of the block and the plane? is there another constraint equation I'm missing?
 
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  • #2
Welcome to PF!

Hi bon! Welcome to PF! :wink:
bon said:
is there another constraint equation I'm missing?

The geometrical constraint is that the block stays on the plane …

how would you write that? :smile:

(oh, and I'd go for conservation of energy)
 
  • #3


tiny-tim said:
Hi bon! Welcome to PF! :wink:


The geometrical constraint is that the block stays on the plane …

how would you write that? :smile:

(oh, and I'd go for conservation of energy)

ahh thanks i think I've got it now..

just working it through..stuck on the last line..

the bottom of the fraction should be sin^2(theta) + M/m

but i have M/m + M/m (sin^2(theta)) - sin^2(theta)...

i know these are equivalent..but how do i simplify my expression to sin^2(theta) + M/m
 
  • #4
actually i think I've made a mistake..

i can't see why though in my working!

the constraint is: tan theta = ay (block) / ax (block) + ax (plane), yes?
 
  • #5
Hi bon! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
bon said:
the constraint is: tan theta = ay (block) / ax (block) + ax (plane), yes?

Nooo …

try it with x and y, rather than ax and ay,

and concentrate on the position of the block relative to a fixed point on the plane. :wink:
 
  • #6
thanks.. :)

so so tanθ = y / xblock - xplane

or tanθ = y / xplane - xblock??
which one?

also then could you differentiate twice to get ratio of accel's? what would the ratio be?

Many thanks again
 
  • #7
bon said:
so tanθ = y / xblock - xplane

or tanθ = y / xplane - xblock??
which one?

(hmm … brackets would help :redface:)

you decide! :smile:
could you differentiate twice to get ratio of accel's? what would the ratio be?

A linear combination of x and y will be the same linear combination of ax and ay :wink:
 
  • #8
tiny-tim said:
(hmm … brackets would help :redface:)

you decide! :smile:


A linear combination of x and y will be the same linear combination of ax and ay :wink:


i really don't know which one it is?! how can i decide?

thanks
 
  • #9
Such problems can be solved by identifying the forces acting on each objects, and noting down their relations.
In this problem, the forces acting on the block are
mg*sinθ..along the inclined plane
mg*cosθ ...perpendicular to the inclined plane
Normal reaction N due to the wedge. Since wedge is moving horizontally, mgcosθ is not equal to N. The block is moving along the wedge.
So (mg*cosθ - N) = m*ay, where ay is the acceleration of m along the normal,say y.
Hence N = (mg*cosθ - m*ay)...(1)
When M moves horizontally through X, block moves through y along normal such that
y = X*sinθ. Οr ay = Ax*sinθ ..(2) where Ax is the acceleration of the wedge.
Forces acting on wedge are
Weight Mg downward, normal reaction N1 from the ground upward and normal force N due to block. Only component which moves the wedge horizontally is N*sinθ
So N*sinθ = M*Ax ...(3)
Using equations (1) and (2) solve for Ax.
 

FAQ: How to Find the Acceleration of a Block and an Inclined Plane?

What is a hard inclined plane problem?

A hard inclined plane problem is a type of physics problem that involves calculating the forces and motion of an object on an inclined plane, taking into account factors such as friction, gravity, and the angle of the incline.

How do you solve a hard inclined plane problem?

To solve a hard inclined plane problem, you first need to identify and draw all the forces acting on the object, including the weight, normal force, and friction. Then, you can use trigonometry and Newton's laws of motion to calculate the acceleration, velocity, and displacement of the object.

What is the difference between a hard and an easy inclined plane problem?

A hard inclined plane problem typically involves more complex factors, such as non-zero friction or a variable angle of incline, while an easy inclined plane problem often assumes ideal conditions with no friction and a constant angle of incline.

How can I determine the coefficient of friction in a hard inclined plane problem?

The coefficient of friction can be determined experimentally by measuring the force required to move an object on an inclined plane and using the formula μ = Ff/FN, where μ is the coefficient of friction, Ff is the force of friction, and FN is the normal force.

What are some real-world applications of hard inclined plane problems?

Hard inclined plane problems are commonly used in engineering and physics to analyze the motion of objects on ramps, hills, and other inclined surfaces. They are also relevant in everyday situations, such as pushing a heavy object up a ramp or calculating the angle of a wheelchair ramp for accessibility.

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