You pull on a crate using a rope as in (Figure 1), except the rope is at an angle of 20.0 ∘ above the horizontal. The weight of the crate is 245 N , and the coefficient of kinetic friction between the crate and the floor is 0.270. What must be the tension in the rope to make the crate move at a constant velocity?
Express your answer with the appropriate units.

What is the normal force that the floor exerts on the crate?
Express your answer with the appropriate units.

The force due to friction opposes the force due to gravity which would cause the object to slide down. (The force due to friction acts up the incline). Additionally, the force due to the rope is also upward.

Let up the incline be positive, and down the incline be negative.

Doing a summation of forces:
[tex]\Sigma F = F_T + F_f - F_g[/tex]

For the crate to be moving at a constant velocity, there is NO net force acting on the crate, so:
[tex]0 = F_T + F_f - F_g[/tex]

Now, we can express each force as an equation.

Force due to tension:

Must be solved for.

Force due to gravity:

On an incline, this is equivalent to the SINE component of its weight. (Force of gravity is STILL THE WEIGHT, but on an incline, it contains a horizontal component that contributes to the net force)

This is expressed as:
[tex]F_g = Mgsin\theta[/tex]

Force due to friction:

Equivalent to the normal force and coefficient of friction. The normal force is the VERTICAL component of the object's weight, so:

[tex]N = Mgcos\theta[/tex]

[tex]F_f = \mu Mgcos\theta[/tex]

Now, plug these expressions into the above equation.

[tex]0 = F_T + \mu Mgcos\theta - Mgsin\theta\\\\Mgsin\theta - \mu Mgcos\theta = F_T[/tex]

Mg = 245 N (weight). Plug in all values:
[tex]245sin(20) - 0.270(245)cos(20) = F_T\\\\F_T = \boxed{21.634 N}[/tex]

The normal force is equivalent to the vertical component (PERPENDICULAR TO THE INCLINE) of the weight (cosine), so:

[tex]N = Mgcos\theta\\\\N = 245cos(20) = \boxed{230.225 N}[/tex]