You pull a suitcase along the floor by exerting 43N at an angle. The force of friction is 27 N and the suitcase moves at a constant speed. What angle does the handle make with the horizontal?

To find the angle at which the handle of the suitcase is being pulled with respect to the horizontal, we need to understand the forces acting on the suitcase. Given that the suitcase is moving at a constant speed, the net force acting on the suitcase must be zero. This means that the horizontal component of the pulling force must be equal to the force of friction.

The force exerted on the handle can be broken down into two components: a horizontal component (parallel to the floor) and a vertical component (perpendicular to the floor). Let's denote the force exerted on the handle as \( F \), the force of friction as \( f_{friction} \), and the angle at which the handle is pulled as \( \theta \).

We can use trigonometry to find the horizontal component of the pulling force, which is given by: \( F_{horizontal} = F \cdot \cos(\theta) \)

Since the suitcase moves at a constant speed, there is no net force in the horizontal direction, which means \( F_{horizontal} \) must be equal to the force of friction: \( F_{horizontal} = f_{friction} \)

So we have: \( F \cdot \cos(\theta) = f_{friction} \)

Now we can solve for \( \theta \): \( \cos(\theta) = \frac{f_{friction}}{F} \) \( \theta = \cos^{-1}\left(\frac{f_{friction}}{F}\right) \)

Plugging in the values we have: \( F = 43 \, N \) \( f_{friction} = 27 \, N \)

\( \theta = \cos^{-1}\left(\frac{27}{43}\right) \)

Now you can use a calculator to find the angle \( \theta \): \( \theta = \cos^{-1}\left(\frac{27}{43}\right) \approx 51.6^\circ \)

So the angle the handle makes with the horizontal is approximately 51.6 degrees.