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27 December, 00:56

A large crate is at rest on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.500. A force F⃗ is applied to the crate in a direction 30.0∘ above the horizontal. The minimum value of F required to get the crate to start sliding is 410 N.

Part A

What is the mass of the crate?

Express your answer with the appropriate units.

+4
Answers (2)
  1. 27 December, 01:05
    0
    93.38 kg

    Explanation:

    We assume that "30.0° above the horizontal" means the force effectively provides lift as well as a sideways force.

    The sideways force will be ...

    410 N * cos (30°) ≈ 355.07 N

    Since the coefficient of friction is 0.500, the vertical force due to gravity, and partially counteracted by the applied force, must be ...

    355.07/0.500 = 710.14 N

    The upward force applied to the crate is ...

    410 N * sin (30°) = 205 N

    so the force due to gravity must be ...

    710.14 N + 205 N = 915.14 N

    This is the product of mass and acceleration. If we assume the acceleration due to gravity is 9.8 m/s², then the mass must be ...

    m = F/a = 915.14 N / (9.8 m/s²) = 93.38 kg

    The mass of the crate is about 93.38 kilograms.
  2. 27 December, 01:24
    0
    93.4 kg

    Explanation:

    Draw a free body diagram. There are three four forces:

    Weight force mg pulling down,

    Normal force N pushing up,

    Friction force Nμ pushing left,

    Applied force F pulling up and to the right, 30.0° above the horizontal.

    Sum of forces in the y direction:

    ∑F = ma

    N + F sin 30.0° - mg = 0

    N = mg - ½ F

    Sum of forces in the x direction:

    ∑F = ma

    F cos 30.0° - Nμ = 0

    ½√3 F = Nμ

    Substitute:

    ½√3 F = (mg - ½ F) μ

    ½√3 F / μ = mg - ½ F

    ½√3 F / μ + ½ F = mg

    ½F (√3 / μ + 1) = mg

    m = F (√3 / μ + 1) / (2g)

    Plug in values:

    m = 410 N (√3 / 0.500 + 1) / (2 * 9.8 m/s²)

    m = 93.4 kg
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