# What is the acceleration of the given wedge?

Hi could anyone help me see my error, thanks.

There is a diagram so I've attached the question as an image:

Picture of question

Here's my attempt:

Resolving forces vertically:

$100g + 6g\,cos(30) = N$

and as we are given $F=\frac{N}{7}$ so we have

$F=\frac{g}{7}(100+3\sqrt(3))$

Resolving forces horizontally:

Resultant horizontal force is

$F - 6g\,sin(30) = F - 3g$

Using $F=ma$ gives:

$\frac{g}{7}(100+3\sqrt(3))-3g=10a$

and so

$a=\frac{g}{70}(79+3\sqrt(3))$

$a=\frac{g}{70}(11-3\sqrt(3))$

Thanks for any help, Mitch.

• Are those units g as in gravitational acceleration or grams? Jan 2 '18 at 17:17
• g is for gravitational acceleration Jan 2 '18 at 17:18
• Where is that $100$ coming from, the weight should be $10g$ Jan 2 '18 at 17:34
• Thanks. I read and re-read and missed that typo each time. Jan 3 '18 at 9:37

Your $F$ is nearly correct, but you seem to have written $100$ instead of $10$,
i.e. $F= \frac{g}{7}(10+3\sqrt{3})$.
$10a=6g\sin(30)-\frac{g}{7}(10+3\sqrt{3})$