A50 kg box sits at rest on a 30 degree ramp where the coef of static friction is 0.5773. if your push was directed at an angle of 40 degrees to the ramp, how hard would you have to push to get the box to move up the ramp?

Given data:

m= 50 Kg,

W= m×g = 50 × 9.81 = 490.5 N

ramp angle (α) = 30 degrees,

coefficient of friction (μs) = 0.5773,

Push at an angle (Θ) = 40 degrees,

Determine: Push to get box move up (P)=?

From the figure,

Resolving the forces along the plane

W sinα + μs.R = P cos Θ       --------------------- (i)

Resolving the forces perpendicular to the inclined plane

W cosα = R+Psin Θ  =>  R= W cosα - Psin Θ -------------- (ii)

Solving (i) and (ii) and keeping μs = tan Φ, Φ = Θ

Pmin = W sin( α +Θ  )

         = W[ sin α.Cos Θ + cos α.sin Θ]

          = 490.5 [ (sin 30.cos40) + (cos30.sin 40)]

          = 460.9 N

Minimum push required to move the box up the ramp is 460.9 N

explanation: magnetic pole, region at each end of a magnet where the external magnetic field is strongest. the south-seeking pole, or any pole similar to it, is called a south magnetic pole. unlike poles of different magnets attract each other; like poles repel each other.