Question

A 2.6 kg ball attached to a string rests on the floor. If the ball's rope is pulled up at a 39-degree angle with a force of 8.2 N, what is the approximate magnitude of the normal force (in N) exerted on the ball by the floor?

Answer 1

To determine the magnitude of the normal force exerted on the ball by the floor, we need to consider the forces acting on the ball.

In this scenario, the ball has two vertical forces acting on it: its weight (mg) directed downward and the normal force (N) exerted by the floor directed upward. The tension in the rope is acting at an angle of 39 degrees with the vertical.

The weight of the ball can be calculated using the formula:

Weight = mass * gravity

where the mass is given as 2.6 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Weight = 2.6 kg * 9.8 m/s^2 = 25.48 N

Since the ball is at rest, the vertical forces must be in equilibrium. This means that the normal force must balance out the weight of the ball.

Therefore, the magnitude of the normal force exerted on the ball by the floor is equal to the weight of the ball:

Magnitude of the normal force = Weight = 25.48 N

Hence, the approximate magnitude of the normal force exerted on the ball by the floor is 25.48 N.

Answer 2

The approximate magnitude of the normal force exerted on the ball by the floor is approximately 30.54 N.

Step-by-step explanation:

To find the approximate magnitude of the normal force exerted on the ball by the floor, we need to consider the forces acting on the ball. Firstly, we have the weight of the ball acting downwards, which can be calculated using the equation Weight = mass * gravity. In this case, the weight is (2.6 kg) * (9.8 m/s^2) = 25.48 N. Secondly, we have the force applied by pulling the rope at a 39-degree angle. We can find the vertical component of this force using the equation Force_vertical = Force_applied * sin(angle). In this case, the vertical component is (8.2 N) * sin(39°) ≈ 5.06 N. Therefore, the approximate magnitude of the normal force is the sum of the weight and the vertical component of the applied force, which is approximately 25.48 N + 5.06 N ≈ 30.54 N.