A 5.7 kg bag of groceries is in equilibrium on an incline of angle θ = 16°. Find the magnitude of the normal force on the bag.
Fn = mg*cos16 = 5.7*9.8*cos16 = 53.7 N.
To find the magnitude of the normal force on the bag, we need to analyze the forces acting on it.
In this scenario, there are three main forces acting on the bag: the weight (mg) acting vertically downwards, the normal force (N) acting perpendicular to the incline, and the component of the weight (mg sinθ) acting parallel to the incline.
Since the bag is in equilibrium, the forces in the vertical direction and the forces in the horizontal direction must balance out.
Let's first consider the forces in the vertical direction:
1. The weight force (mg) acts vertically downwards.
2. The normal force (N) acts perpendicular to the incline, which means it acts in the vertical direction as well.
Since the bag is in equilibrium, the normal force must balance out the weight force. Therefore, the magnitude of the normal force (N) is equal to the weight force (mg).
Here's how to find the magnitude of the normal force:
1. Determine the weight force (W) acting on the bag. The weight force is calculated as the mass (m) of the bag multiplied by the acceleration due to gravity (g). In this case, the weight force is W = mg = (5.7 kg) * (9.8 m/s^2) = 55.86 N.
2. The magnitude of the normal force (N) is equal to the weight force (W). So the magnitude of the normal force is N = 55.86 N.
Therefore, the magnitude of the normal force on the bag is 55.86 N.