How Large Must the Force Be to Keep the Crate Moving at Constant Speed?

  • Thread starter Thread starter HeavOnEarth
  • Start date Start date
  • Tags Tags
    Force Newton
Click For Summary
SUMMARY

The discussion focuses on calculating the force required to keep a crate of books moving at constant speed on a slope of 31 degrees. The crate has a mass of 70 kg and a coefficient of sliding friction of 0.4. The correct force was determined to be 0.905 kN after accounting for gravitational and frictional forces. The user initially miscalculated the force due to incorrect component analysis but later clarified the correct approach to solving the problem.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Knowledge of friction coefficients and their application
  • Ability to resolve forces into components
  • Familiarity with basic trigonometric functions (sine and cosine)
NEXT STEPS
  • Study the concept of force components in physics problems
  • Learn about frictional forces and their calculations in inclined planes
  • Practice solving problems involving forces on slopes
  • Review AP Physics C exam preparation materials and practice problems
USEFUL FOR

Students preparing for the AP Physics C exam, physics tutors, and anyone interested in mastering force calculations in mechanics.

HeavOnEarth
Messages
1
Reaction score
0

Homework Statement



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.

Homework Equations


Fnet=ma
u*F= Ff

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 I am 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~
i got the answer
for anyone who's interested here's 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 kNi 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:
Physics news on Phys.org
Welcome to PF!

HeavOnEarth said:
i would still appeiciate any form of physics whatsoever, tips and help drawing digrams, overall picture, mindset, some practice problems - recommend books/ websites

Welcome to PF! :smile:

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 … " :smile: