A wooden crate rests on a hard wood floor. The static coefficient of friction is 0.5.?

Question

If the crate has a mass of 100kg how much force would be rquired to start it sliding...
I want to kno the formula. am not sure the 1 am using is correct

Answer 1

The applied force has to be greater than the force of friction resisting the motion of the crate. Let F be the force. Then

F - umg > 0 --> F = umg = 0.5*100kg *9.8 m/s^2 = 490 Nts

any force larger than this will start moving the crate.

Answer 2

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Answer 3

I suggest to apply the force F at angle α to the horizon:

------- ------- ------- ------- ------- rope
------- ------- ------- ------- rope
------- ------- ------- rope
------- ------- rope
crate crate
crate crate
crate crate

Friction = μN = μ(mg - Fsinα) = Fcosα
μmg = Fcosα + μFsinα
μmg = F(cosα + μ sinα)
μmg = F √(1 + μ²) cos(α - arctan(μ))


Answer: Fmin = μmg/√(1 + μ²) = 439 N
applied at angle arctan(μ) = 26 degrees