Newton's Force Problem?

Question

A box with a mass of 20 kg, slides a ramp with an acceleration of 0.4 m/s2. The ramp is at an angle θ = 10°. What is the force of friction and coefficient of dynamic friction, μk, between the box and the ramp.
(I don't understand non-equilibrium questions)

Answer 1

Find the force with no friction and the force required for .4m/s².
Subtract them and you have the friction force:

Fnf = Mg*sin10°
Fwf = M*a = .4M

Ff = Fnf - Fwf = Mgsin10°- .4M = M*(gsin10° - .4) = 20(9.8sin10°-.4) = 26.03 N

µ = Ff/Fn = 26.03/Mgcos10° = 26.03/(20*9.8*cos10°) = .135

Answer 2

A non-equilibrium situation is always governed by the equation
F=m*a, where F is the net force. If the net force is zero, the acceleration is zero. If the net force is non zero, acceleration is non zero. Also, F and a are vectors.

Okay. Consider a free body diagram of the box. There is a force m*g pulling straight down on the box. If we translate our coordinate system such that x is parallel to the surface of the incline, then the force has an x component of
m*g*sin(θ) and a y component that is m*g*cos(θ).

The other force acting on the block is friction. It is parallel to the surface of the incline and opposite to the direction of motion. Since we are dealing with vectors, only the forces in the x are related to the acceleration of the box down the ramp. F=m*g*sin(θ) -N*μk, where N is the normal force.

The normal force is the force perpendicular to the interface of the box and the incline, which we identified above as
m*g*cos(θ).
so, the net Forces acting on the block in the x direction in our new coordinate system are
F=m*g*sin(θ)-m*g*cos(θ)*μk

The y direction doesn't matter anymore since there is no acceleration in that direction in our coordinate system.
Also F=m*a
so
m*g*sin(θ)-m*g*cos(θ)*μk=m*a
now divide out m, simplify a bit, and we have an expression for a
a=g*(sin(θ)-cos(θ)*μk)

While the box accelerates, this a is constant
In fact, it was given as 0.4 m/s^2 and θ = 10°,
so plug them in and solve for μk.

j

Answer 3

a) The wagon movements at consistent p.c... hence, thepersistent of the pulling is same with the frictionalpersistent. F = ma consistent p.c.., F cos 30 = 30 F = 30 / (cos 30) F = 34.641 N b) Now the wagon quickens at 0.4 m/s^2, there is an unbalancedpersistent. it particularly is once you talk that thepersistent of pulling is larger than frictionalpersistent. subsequently, tension of pulling - Frictionalpersistent = Unbalanced tension. tension of pulling = Unbalancedpersistent + Frictional tension tension of pulling = 20 kg (0.4 m/s^2) + 30 N tension of pulling = 38 N tension of pulling = F cos 30 = 38 F = 38 / (cos 30) F = 40 3.87 N