"A 25-N crate is held at rest on a frictionless incline by a force that is parallel to the incline. If the incline is 25◦ above the horizontal the magnitude of the normal force of the incline on the crate is: "?
Well, it seems like the crate is really feeling the pressure to stay put on that incline! But not to worry, because the normal force has got its back. In this case, since the incline is at an angle of 25 degrees above the horizontal, the magnitude of the normal force would be equal to the weight of the crate, which is 25 Newtons. So, you could say the crate is getting a "normal" amount of support from the incline. Keep up the good work, Mr. Crate!
To find the magnitude of the normal force acting on the crate on the incline, we need to consider the forces acting on the crate.
1. Gravity force: The weight of the crate acts straight downward and can be calculated using the formula:
Weight = mass * gravity
Given that the weight of the crate is 25 N, we can calculate the mass by rearranging the formula:
mass = weight / gravity
Assuming gravity is approximately 9.8 m/s^2, the mass of the crate is:
mass = 25 N / 9.8 m/s^2
2. Normal force: The normal force acts perpendicular to the incline and counters the force of gravity. Since the crate is at rest, the normal force is equal in magnitude but opposite in direction to the weight. Therefore, the normal force is 25 N.
Note: The angle of inclination does not affect the magnitude of the normal force.
To find the magnitude of the normal force of the incline on the crate, we need to understand the forces acting on the crate.
In this scenario, we have a crate on an incline, and the force holding the crate at rest is parallel to the incline. We can break down the forces acting on the crate into two components - one parallel to the incline and one perpendicular to the incline.
1. The weight of the crate, which is the force acting vertically downward. We can find this by multiplying the mass of the crate by the acceleration due to gravity (g ≈ 9.8 m/s^2).
Weight (W) = mass (m) * acceleration due to gravity (g)
2. The force holding the crate at rest, which is parallel to the incline. This force is given as 25 N in the question.
Now, let's find the normal force, which is the perpendicular force exerted by the incline on the crate.
The force components can be related to each other by using trigonometry, as follows:
Weight (W) = mg * sin(θ)
Force parallel to incline (F_parallel) = mg * cos(θ)
where θ is the angle of the incline (25 degrees in this case).
Since the crate is at rest, the force parallel to the incline cancels out the force of gravity along the incline.
F_parallel = Weight (W)
Now, we can find the normal force (N) using the force perpendicular to the incline:
Normal force (N) = Weight (W) * cos(θ)
Substituting the values:
Normal force (N) = (mg * sin(θ)) * cos(θ)
We know that the weight (W) equals mg, so we can simplify the equation:
Normal force (N) = mg * sin(θ) * cos(θ)
Plugging in the values:
Normal force (N) = (25 kg * 9.8 m/s^2) * sin(25°) * cos(25°)
Evaluating this equation will give you the magnitude of the normal force acting on the crate.