A 20 kg bag of luggage is rolled across a horizontal floor by a force of 50 N applied to the handle which makes an angle of 65 degrees with the horizontal. If the bag moves at constant speed, what is the coefficient of friction between the bag and the floor?
To find the coefficient of friction between the bag and the floor, we first need to identify the forces acting on the bag.
1. The applied force: This is the force exerted by the person rolling the bag, which has a magnitude of 50 N and an angle of 65 degrees with the horizontal. Let's decompose this force into its horizontal and vertical components.
Horizontal component = 50 N * cos(65°)
Vertical component = 50 N * sin(65°)
2. The normal force (N): This is the force exerted by the floor on the bag perpendicular to the contact surface.
3. The gravitational force (mg): This is the weight of the bag, which is given as 20 kg. The magnitude of the gravitational force is calculated as:
Gravitational force = mass * acceleration due to gravity
Gravitational force = 20 kg * 9.8 m/s^2
Now, since the bag is moving at a constant speed, we know that the net force acting on it is zero. In other words, the horizontal component of the applied force must balance out the force of friction.
Let's set up the equation:
Net force = 0
Horizontal component of applied force - Force of friction = 0
Now, let's find the horizontal component of the applied force:
Horizontal component = 50 N * cos(65°)
Next, let's substitute the expression for the horizontal component into the equation:
50 N * cos(65°) - Force of friction = 0
Now, rearrange the equation to solve for the force of friction:
Force of friction = 50 N * cos(65°)
Finally, we calculate the coefficient of friction by dividing the force of friction by the normal force:
Coefficient of friction = Force of friction / (mass * acceleration due to gravity)
Substituting the values:
Coefficient of friction = (50 N * cos(65°)) / (20 kg * 9.8 m/s^2)
Calculating this expression gives us the coefficient of friction between the bag and the floor.