A woman at an airport is towing her 17.0 kg suitcase at constant speed by pulling on the strap at an angle of θ above the horizontal (Fig. P4.40). She pulls on the strap with a 26.0 N force, and the friction force on the suitcase is 10.0 N. What angle does the strap make with the horizontal? What normal force does the ground exert on the suitcase?

To find the angle θ that the strap makes with the horizontal, we can use trigonometry and the given forces.

First, let's resolve the forces acting on the suitcase in the horizontal and vertical directions.

In the horizontal direction:
- The applied force Fapplied is acting in this direction.

In the vertical direction:
- The weight force Fweight is acting downwards.
- The normal force Fnormal is acting upwards.
- The friction force Ffriction is acting upwards (opposite in direction to Fapplied).

Since the suitcase is moving at constant speed, the net force in both the horizontal and vertical directions must be zero.

Now, let's break down the given information:
- Fapplied = 26.0 N
- Ffriction = 10.0 N

To find the weight force, we can use the equation:
Fweight = mg
where m is the mass of the suitcase and g is the acceleration due to gravity.

Given:
m = 17.0 kg (mass of the suitcase)

Let's calculate the weight force:
Fweight = 17.0 kg * 9.8 m/s^2 (approximate value of g)
Fweight = 166.6 N (rounded to one decimal place)

Now, let's analyze the vertical forces:
In the vertical direction, the equation becomes:
ΣFvertical = Fnormal + Fweight - Ffriction = 0

Substituting the known values:
Fnormal + 166.6 N - 10.0 N = 0

Solving for Fnormal:
Fnormal = 166.6 N - 10.0 N
Fnormal = 156.6 N

The normal force exerted by the ground on the suitcase is 156.6 N.

Now, to find the angle θ that the strap makes with the horizontal, we can use the trigonometric relationship between the horizontal and vertical forces.

The relationship is given by:
tan(θ) = (Fweight - Ffriction) / Fapplied

Substituting the known values:
tan(θ) = (166.6 N - 10.0 N) / 26.0 N

Solving for θ:
θ = tan^(-1) ((166.6 N - 10.0 N) / 26.0 N)

Using a calculator, we find:
θ ≈ 80.7 degrees

Therefore, the angle that the strap makes with the horizontal is approximately 80.7 degrees.