A box of 65.0 kg is going down with a 37.2 degree. It has a mu of 0.107 , what is N?
回答 (2)
2018-01-06 10:24 pm
Take g = 9.81 m/s²
Component of gravitational force going down the incline = 65.0 × 9.81 × sin37.2° N
Frictional force acting on the box = (65.0 × 9.81 × cos37.2°) × 0.107 N
Net force acting on the box, F = [65.0 × 9.81 × sin37.2° - (65.0 × 9.81 × cos37.2°) × 0.107] N = 331 N
Component of gravitational force going down the incline = 65.0 × 9.81 × sin37.2° N
Frictional force acting on the box = (65.0 × 9.81 × cos37.2°) × 0.107 N
Net force acting on the box, F = [65.0 × 9.81 × sin37.2° - (65.0 × 9.81 × cos37.2°) × 0.107] N = 331 N
2018-01-06 8:48 pm
Next time, try to copy the question accurately.
Anyway, in your problem, the normal force is
(65.0 kg)cos(37.2 deg)(9.8 m/s^2) = about 500N but use a calculator.
Anyway, in your problem, the normal force is
(65.0 kg)cos(37.2 deg)(9.8 m/s^2) = about 500N but use a calculator.
收錄日期: 2021-04-18 18:01:48
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