Comments on: What is the required force to pull a shopping car?

By: Matt

Matt — Tue, 03 Nov 2009 02:35:54 +0000

for the diagram—
[only ten people can dwnld]
[some one please re-upload it after reading the answer]

this is the hint
if u are able to solve it then exellent.
if not please ask again. Matt

By: Questor

Questor — Sun, 01 Nov 2009 07:37:15 +0000

……..Fy………F
…W…↑…….⁄
….↓….|….⁄_35°
._▓▓._|.⁄——→Fx
╘O=O╛←――Ff
▒▒↑▒▒▒▒▒▒▒▒▒▒▒▒
…..|
….N.

Given:
W = mg = 20kg(9.8)
µ = 0.05

(A) Force needed to PULL the car at a constant speed.
‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾
Summation of forces along the Y-axis = 0
∑Fy = 0
0 = -W + N + Fy
…= -mg + N + Fsin35°
N = mg - Fsin35° ………..◄eq1

Summation of forces along the X-axis = 0
∑Fx = 0
0 = Fx - Ff
Ff = Fx
Ff = Fcos35°

But Ff = µN … Questor

By: Itzel

Itzel — Thu, 29 Oct 2009 02:01:53 +0000

The formula for friction is

Fr = μR

Where μ is the coefficient of friction, and
R is the normal contact force.

So the normal contact force will be
Mass x gravity = 50*9.81 = 490.5

0.37 * 490.5 = 181.485 N

It is worth noting that that answere is too precise and also that I ignored the vectors and just used magnitude, in more difficult problems you will need to use vectors. Itzel