A 4.5 kg bag of groceries is in equilibrium on an incline of angle θ = 55°. Find the normal force of the incline on the bag.
M g cos55
M = 4.5 kg
g = 9.8 m/s^2
The normal force will be in Newtons
To find the normal force of the incline on the bag, we can use Newton's second law and resolve the forces acting on the bag.
The weight of the bag is given by the formula:
Weight = mass * acceleration due to gravity
In this case, the mass of the bag is 4.5 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
First, we need to find the force of gravity acting on the bag. The formula for the force of gravity is:
Force of gravity = mass * acceleration due to gravity
So, the force of gravity acting on the bag is:
Force of gravity = 4.5 kg * 9.8 m/s^2
To find the normal force, we need to consider the forces acting on the bag along the vertical axis. The weight of the bag acts vertically downward.
The normal force acts perpendicular to the incline and opposes the weight of the bag. Therefore, the normal force is equal in magnitude but opposite in direction to the force of gravity.
Since the incline is at an angle of 55 degrees with the horizontal, we can resolve the force of gravity into two components: one parallel to the incline and the other perpendicular to the incline.
The component of the weight parallel to the incline is given by:
Force parallel = Force of gravity * sin(θ)
And the component of the weight perpendicular to the incline is given by:
Force perpendicular = Force of gravity * cos(θ)
Since the normal force opposes the force of gravity perpendicular to the incline, the normal force is equal in magnitude but opposite in direction to the force perpendicular:
Normal force = -Force perpendicular
Therefore, the normal force of the incline on the bag is:
Normal force = -Force of gravity * cos(θ)