# Dynamics question

## Homework Statement

a mass z is held on an incline plane with mass P. when the the masses are released, both objects begin to move and accelerate. find accell of the incline plane

GIVEN: z,P,angle A, surfaces are frictionless

solve in terms of these variables.

f=ma

## The Attempt at a Solution

i chose my positive x direction along the ground surface below the incline plane, toward the direction of its motion.

so first i solved for the small mass on top the normal force, which is the force which applied onto the incline plane by the small mass.

fnety=0=FNcosA - zg
FN= zg/cosA

now when i do the FBD of the incline plane the forces acting on it, are the normal (FN) which gives it an unbalanced force in the +X direction.

so i get the expression

fnetx= (z+P)a=FNsinA

subbing in the FN,i get

a= zgtanA/(z+P)

is this correct? if not can you tell me where i am going wrong.

Last edited:

nasu
Gold Member
The small mass accelerates along the y direction. So Fnet is not zero along this direction.

Ahh I see..
Fnety=zg=FN-zgcosA
FN=2zg/cosA

Ans; 2zgtanA/(z+P)

nasu
Gold Member
The vertical acceleration of the small object is not g. It's not falling freely.

The vertical acceleration of the small object is not g. It's not falling freely.

would it make it more simple if i chose a different frame of reference.

nasu
Gold Member
You could work in the non-inertial frame of the moving inclined plane. But I am not sure it makes simpler. Depends how it is easier to visualize for you. In the inertial frame attached to the ground or in the moving frame.