## What is the required force to pull a shopping car?

A woman pulls a shopping car with a force F at an angle of 35. The car mass is 20kg and its friction coefficient is 0.05. Find:

(A) The force the woman needs to apply to move the car with a constant velocity of 1m/s.

(B) The magnitude of the normal force and friction when the car moves at constant velocity.

(C) The force the woman needs to exert to push the car with a constant velocity of 1/s.

(D) The Displacement if the car is pushed 5 seconds and the displacement if the car is pulled 5 seconds.

July 26th, 2010 at 6:47 pm

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.

July 26th, 2010 at 7:45 pm

……..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 …<==substitute value of N from ◄eq1 Fcos35° = µ [ mg - Fsin35°] F(.819) = 0.05[20(9.80) - F(.573)] .............= 9.8 - F(0.029) F(0.848) = 9.8 F = 11.56N (B) Normal force N = mg - Fsin35° ...........◄eq1 ...= 20 (9.80) - 11.56(.573) N = 196 - 6.622 N = 189.38N (C) WHEN PUSHING the shopping car ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾ ........Fy.........F ...W...|.......⁄ ....↓....|....⁄_35° ._▓▓._↓.⁄←------Fx ╘O=O╛――→Ff ▒▒↑▒▒▒▒▒▒▒▒▒▒▒▒ .....| ....N. Summation of forces along the Y-axis = 0 ∑Fy = 0 0 = -W + N - Fy ...= -mg + N Fsin35° N = mg + Fsin35° ...........◄eq1b Summation of forces along the X-axis = 0 ∑Fx = 0 0 = - Fx + Ff Ff = Fx Ff = Fcos35° But Ff = µN ...<==substitute value of N from ◄eq1b Fcos35° = µ [ mg + Fsin35°] F(.819) = 0.05[20(9.80) + F(.573)] .............= 9.8 + F(0.029) F(0.79) = 9.8 F = 12.40 N (D) Displacement whether being pulled or pushed will be the same since there is no Unbalanced Force because the velocity is constant at 1m/sec. Hence there is also no acceleration involved ......._ S = Vt S = (1m/sec)5 S = 5m (the same for pulling or pushing)

July 26th, 2010 at 8:29 pm

for the diagram—

http://rapidshare.com/files/295889546/Untitled.jpg.html

[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.