1

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

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

But the book answer is:

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

Thanks for any help, Mitch.

gnitsuk
  • 217

1 Answers1

0

Your $F$ is nearly correct, but you seem to have written $100$ instead of $10$,

i.e. $F= \frac{g}{7}(10+3\sqrt{3})$.

Since F is your Friction force, and we know that friction acts in the opposite direction to movement, the diagram tells us the object is moving right, so your equation for acceleration should be:

$10a=6g\sin(30)-\frac{g}{7}(10+3\sqrt{3})$

This should simplify to the answer given in the book.

cluelessatthis
  • 465