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May 11th, 2010 • 19:05

What Is The Required Force To Pull A Shopping Car?

A woman pulls a car with a 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.

3 Comments

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  • May 11th 2010 • 21:05
    by Itzel

    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.

  • May 11th 2010 • 23:05
    by Questor

    ……..Fy………F
    …W…↑…….⁄
    ….↓….|….⁄_35°
    ._â–“â–“._|.⁄——→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° ………..â—„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 â—„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° ………..â—„eq1
    …= 20 (9.80) – 11.56(.573)
    N = 196 – 6.622
    N = 189.38N
    (C) WHEN PUSHING the shopping car
    ‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾…
    ……..Fy………F
    …W…|…….⁄
    ….↓….|….⁄_35°
    ._â–“â–“._↓.⁄←——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° ………..â—„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 â—„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)

  • May 11th 2010 • 23:05
    by JerichoX

    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.

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