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27 November, 11:31

A 12-kg crate rests on a horizontal surface and a boy pulls on itwith a force that is 30? below the horizontal. If the coefficient of static friction is 0.40, theminimum magnitude force he needs to start the crate moving is: A. 44N B. 47N C. 54N D. 56N E. 71N

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Answers (2)
  1. 27 November, 11:35
    0
    E. 71 N

    Explanation:

    Given Data

    Mass=12 kg

    Angle=30°

    static friction=0.44 N

    To Find

    Force=?

    Solution

    Sum of the forces in the direction of motion is given as

    = 0 = F*cos (30) - friction

    Sum of the forces in the vertical direction (up is positive) is given as

    = 0 = N - mg - F*sin (30)

    N = m*g + F*sin (30)

    From the definition of friction

    friction = u*N

    friction = u * (m*g+F*sin (30))

    Plug back into the top

    0 = F*cos (30) - (u*m*g+u*F*sin (30))

    Solve for F

    u*m*g = F * (cos (30) - u*sin (30))

    F = (u*m*g) / (cos (30) - u*sin (30))

    Put the number we get

    F = 71 N

    So Option E is correct
  2. 27 November, 11:41
    0
    C. 54N

    Explanation:

    From Friction

    F' = μR ... Equation 1

    Where F' = Friction force, μ = Coefficient of static friction, R = Normal reaction

    But,

    R = mg ... Equation 2

    Where m = mass of the crate, g = acceleration due to gravity of the crate.

    substitute equation 2 into equation 1

    F' = μmg ... Equation 3

    Given: m = 12 kg, μ = 0.4, g = 9.8 m/s²

    Substitute into equation 3

    F' = 12 (0.4) (9.8)

    F' = 47.04 N.

    From the question, the boy pulls with a force of 30° below the horizontal.

    F' = Fcos30°

    Where F = The force needed to start the crate moving

    make F the subject of the equation

    F = F'/cos30

    F = 47.04/cos30

    F = 47.04/0.8660

    F = 54.32 N.

    Hence the right option is C. 54N
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