# Mass in accelerated wedge [closed]

I got this exercise from Kleppner's Introduction to Mechanics:

A 45° wedge is pushed along a table with constant acceleration A. A block of mass m slides without friction on the wedge. Find its acceleration.

I tried solving it, but I can't obtain what appears in the Solutions Manual.

I don't understand why everything is multiplied by sin(theta). If you could help me it'd be awesome. Thanks!

• Have you gone through all the trig identities, to see can you make the equation fit his version? Because, obviously, the rest of it other than the first line is ok. All you need to is sort the first line, as you probably know, sorry. – user108787 Aug 15 '16 at 1:13
• Please show YOUR solution to this problem, explaining your reasoning. – sammy gerbil Aug 15 '16 at 1:46
• This is a mathjax-enabled site. Do use this facility. To have a quick review check this meta MSE post: MathJax basic tutorial and quick reference. – user36790 Aug 15 '16 at 5:20

Accelerations in the $x$ and $y$ directions must be summed:

$x-direction$:

$$A_{block/relative/to/slope} = g*sin{\theta}$$

The x-component of that equals:

$$cos{\theta}*g*sin{\theta}$$

The total acceleration of the block in the x-direction equals:

$$A_{wedge} + cos(\theta)*g*sin(\theta)$$

$y-direction$:

$$-g*sin^2{\theta}$$

Therefore:

$$\vec{A_{total}} = [A_{wedge} + cos(\theta)*g*sin(\theta)]i - [g*sin^2{\theta}]j$$

I think this differs from your solution above. Maybe I erred. Carefully review what I did to make sure I'm not mistaken.

The answer depends on the orientation of the wedge with respect to the direction of motion (speed) and acceleration. If the ramp of the wedge is 'the front side' then I can imagine the massive block may be pushed upwards in case acceleration 'A' is high enough. For one particular value of A, the acceleration of the block is exactly A.