What is minimum amount?

Question

A sled of mass 40 kg is pulled along flat, snow-covered ground. The static friction coefficient is 0.28, and the kinetic friction coefficient is 0.0800. What is minimum amount of force needed to make the block move? What is should the Fpull be reduced to, to keep the sled moving at a constant velocity?

 

How fast would a 15 kg box accelerate down a 25 degrees snowy slope if the coefficient of kinetic friction were 0.15?

 

A 108 N lamp is hung up with two ropes that each make 20 degree angles with the horizontal. What is the tension in the ropes? What is the tension if the angle is 60 degree?

Answer 1

Hey Sun -- "stiction" typically exceeds sliding friction ...

400N at 28% stiction needs a 110N start and an 8% "maintenance" push of 30N :)

 

Answer 2

The minimum amount of force needed to make the block move is equal and opposite to the force of static friction:

 

F = -F_fs = µ_s F_n = µ_s mg = 0.28*40 kg*9.8 m/s² = 110 N

 

The minimum amount of force needed to keep the block moving at a constant velocity is equal and opposite to the force of kinetic friction:

 

F = -F_ks = µ_k F_n = µ_k mg = 0.08*40 kg*9.8 m/s² = 31 N

 

Be sure to pay attention to significant digits!

Answer 3

1)

To start moving, use static coefficient.

Fmin = μmg = 0.28(40)g = 109.8 N

 

To keep it moving at a constant velocity, use kinetic coefficient.

F = μmg = 0.08(40)g = 31.36 N

 

2) Done in another post.

 

3) With symmetry, apply Newton's second law in vertical direction:

2Tsinθ = mg

T = 108/(2sinθ)

 

θ = 20^o, T = 158 N

 

θ = 60^o, T = 62.4 N