- #1
housemartin
- 87
- 0
Hello,
A 45o 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
if i make x' be a horizontal coordinate moving with wedge, then this is in not inertial frame so -mA term adds:
m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA (1)
y coordinate is inertial, and Newton second law is:
m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg (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:
-\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta) (3)
sins theta = 45, tan(45)=1
Solving equations (1) and (2) for N, and dividing one by another, i get:
tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g)
Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:
\frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g
Combining this with equation (3), i get:
\frac{d^2 y}{dt^2} = (A - g)/2
Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.
Homework Statement
A 45o 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
Homework Equations
if i make x' be a horizontal coordinate moving with wedge, then this is in not inertial frame so -mA term adds:
m \frac{d^2 x'}{dt^2} = F_x - mA = N cos(\theta) - mA (1)
y coordinate is inertial, and Newton second law is:
m \frac{d^2 y}{dt^2} = F_y = N sin(\theta) - mg (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:
-\frac{d^2 y}{dt^2} = (\frac{d^2 x}{dt^2}-A)tan(\theta) (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:
tan(\theta)(\frac{d^2 x'}{dt^2}+A) = (\frac{d^2 y}{dt^2} + g)
Since acceleration of x' in not inertial frame plus A is acceleration of block in inertial frame, and tan(45) = 1, i get:
\frac{d^2 x}{dt^2} = \frac{d^2 y}{dt^2} + g
Combining this with equation (3), i get:
\frac{d^2 y}{dt^2} = (A - g)/2
Is my solution correct? The fact that when A>g block is moving up kinda confuses me. Sorry for my english.
Last edited by a moderator:
