- #1

housemartin

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Hello,

A 45

[PLAIN]http://img210.imageshack.us/img210/6958/wedge45.jpg [Broken]

if i make x' be a horizontal coordinate moving with wedge, then this is in not inertial frame so -mA term adds:

[tex]m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA[/tex] (1)

y coordinate is inertial, and Newton second law is:

[tex]m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg [/tex] (2)

The only forces as seen by inertial observer are mg and normal force N.

To associate accelerations of x and y in inertial frame I make X (acceleration A) to be position of wedge and x that of block, then:

[tex] -\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta) [/tex] (3)

sins theta = 45, tan(45)=1

Solving equations (1) and (2) for N, and dividing one by another, i get:

[tex] tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g) [/tex]

Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:

[tex] \frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g [/tex]

Combining this with equation (3), i get:

[tex] \frac{d^2 y}{dt^2} = (A - g)/2 [/tex]

Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.

## Homework Statement

A 45

^{o}wedge is pushed along a table with constant acceleration A. A block of mass*m*slides without friction on the wedge. Find its acceleration in y direction. (Gravity directed down, acceleration due to gravity is g).[PLAIN]http://img210.imageshack.us/img210/6958/wedge45.jpg [Broken]

## Homework Equations

if i make x' be a horizontal coordinate moving with wedge, then this is in not inertial frame so -mA term adds:

[tex]m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA[/tex] (1)

y coordinate is inertial, and Newton second law is:

[tex]m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg [/tex] (2)

The only forces as seen by inertial observer are mg and normal force N.

To associate accelerations of x and y in inertial frame I make X (acceleration A) to be position of wedge and x that of block, then:

[tex] -\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta) [/tex] (3)

sins theta = 45, tan(45)=1

## The Attempt at a Solution

Solving equations (1) and (2) for N, and dividing one by another, i get:

[tex] tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g) [/tex]

Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:

[tex] \frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g [/tex]

Combining this with equation (3), i get:

[tex] \frac{d^2 y}{dt^2} = (A - g)/2 [/tex]

Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.

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