Magnitude of acceleration hwk check

[Ashley1nOnly]

Just like you drew it in your original picture, except you would omit $m \vec{g}$ and $\vec{F}_N$.

Yep and I see why now, thanks so much for not giving me the answers and making me work for it.

Am I right now with my mind of thinking

 

[Herman Trivilino]

X
(F1-f2cos(62))/m =a

Y

(F2sin(62))/m =a

Now how do we combine them

The first equation is actually equal to a_x, the second is equal to a_y.

Do you know how to combine them to find the magnitude of $\vec{a}$?

Alternatively, you could have combined F_x and F_y to find the magnitude of $\vec{F}$ first. And then divided that by m to find the magnitude of $\vec{a}$.

 

[Ashley1nOnly]

Just like you drew it in your original picture, except you would omit $m \vec{g}$ and $\vec{F}_N$.

Then going on with the problem to finish it

I have everything summed up
F(net)=(9-8cos(62))i +(F2sin(62))j
=5.24i + 7.06j
Take the magnitude of it
Sqrt( (5.24)^2+(7.06)^2)
=8.8
Which gave me the next force. Now in order to find the acceleration I know that
F(net)=ma

A= f(net)/m
A=8.8/3.0
A=2.93

 

[Herman Trivilino]

Nice!

 

[Ashley1nOnly]

Nice!

Thanks so much.
WOOT woot