What is the required force to pull a shopping car?


car
JerichoX asked:


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.

This entry was posted on Wednesday, October 28th, 2009 at 7:08 am and is filed under Physics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “What is the required force to pull a shopping car?”

  1. Itzel Says:

    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

  2. Questor Says:

    ……..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

  3. Matt Says:

    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

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