A 20 kg wagon is pulled along the level ground by a rope inclined at 30 deg above horizontal. A frictional force of 30 N opposes the motion, how large is the pulling force if the wagon is moving at constant velocity.
F cos 30 = 30
To find the pulling force, we will use the concept of equilibrium. When an object is in equilibrium, the net force acting on it is zero. In this case, since the wagon is moving at constant velocity, the net force acting on it must be zero.
Let's break down the forces acting on the wagon:
1. Gravitational force (weight): The weight of the wagon can be calculated using the formula: weight = mass * acceleration due to gravity. In this case, the mass of the wagon is 20 kg and the acceleration due to gravity is approximately 9.8 m/s^2. So, the weight of the wagon is 20 kg * 9.8 m/s^2 = 196 N, acting vertically downwards.
2. Normal force: The normal force is the force exerted by the ground on the wagon, perpendicular to the ground. Since the wagon is on level ground and not accelerating vertically, the normal force is equal in magnitude but opposite in direction to the weight of the wagon. So, the normal force is 196 N, acting vertically upwards.
3. Frictional force: The frictional force opposes the motion of the wagon and is given as 30 N in the problem statement.
4. Pulling force (unknown): This is the force applied to the wagon to overcome the frictional force and move it with constant velocity. We need to find the magnitude of this force.
Since the wagon is moving at constant velocity, the net force acting on it is zero. Therefore, we can write the equation:
Net force = Pulling force - Frictional force = 0
So, Pulling force = Frictional force = 30 N.
Therefore, the magnitude of the pulling force needed to keep the wagon moving at constant velocity is 30 N.