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.|||........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 ...%26lt;==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 ...%26lt;==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)|||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.|||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.