Final answer:
The frictional force acting on a wagon pulled at a constant velocity by a 100N force at an angle of 30° above the horizontal is approximately 86.6N, calculated using the horizontal component of the applied force.
Step-by-step explanation:
The student has asked about the frictional force acting on a wagon that is being pulled at a constant velocity by a force at an angle. To solve for the frictional force, we must consider the components of the applied force. The force has two components due to the 30° angle: a horizontal component (Fhorizontal) and a vertical component (Fvertical). Since the wagon is moving at a constant velocity, the net force on the wagon is zero, meaning the horizontal component of the applied force is balancing out the frictional force.
To find the horizontal component of the applied 100N force, we use cosine, since cos(30°) = Fhorizontal / 100N. Hence, Fhorizontal = 100N * cos(30°), which is approximately 86.6N. Because the net force is zero, the frictional force must have the same magnitude as this horizontal component but in the opposite direction. Therefore, the frictional force acting on the wagon is approximately 86.6N.