Question

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.

Answer 1

Answer:

  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.

Answer 2

Answer:

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