# Newton law's/ force related

1. Oct 26, 2008

### HeavOnEarth

1. The problem statement, all variables and given/known data

A crate of books is to be put on a truck with
the help of some planks sloping up at 31◦. The
mass of the crate is 70 kg, and the coefficient
of sliding friction between it and the planks is
0.4. You and your friends push horizontally
with a force ~F .
The acceleration of gravity is 9.81 m/s2 .
Once the crate has started to move, how
large must F be in order to keep the crate
moving at constant speed? Answer in units of
kN.

2. Relevant equations
Fnet=ma
u*F= Ff

3. The attempt at a solution

i did .4 x F = sin(31) x 70 x 9.81, solving for F i get 884.192 N/ 1000
= .884192 kN
however it's wrong
so im confused as how to approach this problem

i tried cos(31) as well
and i've tried these solutions; all of which were rejected ( on the UT website )
10/26/08 8:46 PM Try#1: .353677
10/26/08 9:46 PM Try#2: 141.471
10/26/08 9:46 PM Try#3: .141471
10/26/08 9:47 PM Try#4: .235447
10/26/08 9:48 PM Try#5: .884192
10/26/08 9:49 PM Try#6: 1.47154

also i apologize if i posted in the wrong forum/ had bad formatting

oh nevermind lovely physics forum~
for anyone who's interested heres the work process
(excuse my intangible markings i was doing this on another forum as well)

ma = Fpush - Fg - Ff
Ff = .4*(cos31)*686.7 = 235.4
Fg = 686.7*sin31 = 353.7
a = 0

0 = Fpush - 353.7 - 235.4
Fpush = 589.1 N

edit: Ah crap it's 9.81 ~_~
edit2: ah crap you're pushing horizontally

Fpush = Fhor - Ff
589.1 = Fhor*cos31 - .4*Fhor*sin31
589.1 = 0.651*Fhor

Fhor = 905 N
or .905 kN

i would still appeiciate any form of physics whatsoever, tips and help drawing digrams, overall picture, mindset, some practice problems - recommend books/ websites
the main goal for me is to get a 4,5 on the AP physics C exam

Last edited: Oct 26, 2008
2. Oct 27, 2008

### tiny-tim

Welcome to PF!

Welcome to PF!

Tip: if you were systematic in writing the proof, you wouldn't have missed the F sin31º the first time.

You should always state which direction you're taking components in (and why) … don't just do it in your head, write it down

in this case I can see that you took components in the normal direction, so you should have started by saying:

"there is no acceleration in the normal direction, so the sum of the components of force in that direction is zero … "