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
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.
