A 50 kg crate is pushed at constant speed up the frictionless 31° ramp shown in the figure.
A)What horizontal force F is required?
B)What force is exerted by the ramp on the crate?
The component of the horizontal force up the ramp must equal the component of the weight down the ramp, in a frictionless situation with no acceleration.
To solve this problem, we need to consider the forces acting on the crate on the ramp.
First, let's draw a free body diagram for the crate:
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F | |
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Now, let's break down the forces acting on the crate:
1. Weight (mg): The weight of the crate is given by the formula weight = mass * gravitational acceleration. In this case, the mass is 50 kg and the gravitational acceleration is approximately 9.8 m/s^2. So, the weight of the crate is 50 kg * 9.8 m/s^2 = 490 N. This force acts vertically downward.
2. Normal force (N): The normal force is the force exerted by the ramp on the crate perpendicular to the surface of the ramp. Since the crate is at a constant speed, the vertical components of all forces acting on the crate must add up to zero. Therefore, the normal force is equal in magnitude and opposite in direction to the vertical component of the weight. The vertical component of the weight is given by weight * sin(θ), where θ is the angle of the ramp. In this case, θ is 31°. So, the normal force is 490 N * sin(31°) = 254 N. This force acts perpendicular to the ramp, pointing away from the ramp.
3. Force applied horizontally (F): The horizontal force required to push the crate up the ramp at a constant speed is equal in magnitude and opposite in direction to the horizontal component of the weight. The horizontal component of the weight is given by weight * cos(θ). In this case, θ is 31°. So, the horizontal component of the weight is 490 N * cos(31°) = 418 N. Therefore, the horizontal force required (F) is 418 N. This force acts horizontally, parallel to the surface of the ramp.
4. Force exerted by the ramp on the crate: The force exerted by the ramp on the crate is equal in magnitude and opposite in direction to the horizontal force required (F). So, the force exerted by the ramp on the crate is 418 N. This force acts horizontally, parallel to the surface of the ramp.
In summary:
A) The horizontal force required (F) is 418 N.
B) The force exerted by the ramp on the crate is also 418 N.
To find the required horizontal force F and the force exerted by the ramp on the crate, we can use the concepts of forces and Newton's laws of motion.
A) To determine the horizontal force F required to push the crate up the ramp at constant speed, we need to consider the forces acting on the crate. The forces involved are the gravitational force (mg) acting vertically downwards and the force parallel to the ramp (F) acting horizontally.
Since the crate is moving at a constant speed, we know that the net force acting on it is zero. This means that the force of gravity acting downwards must be balanced by the horizontal force F.
Using trigonometry, we can decompose the gravitational force mg into two components: one perpendicular to the ramp (mg * cosθ) and one parallel to the ramp (mg * sinθ). In this case, θ refers to the angle of the ramp, which is given as 31°.
In the vertical direction (perpendicular to the ramp), the normal force of the ramp balances the vertical component of the gravitational force, so there is no net force or acceleration in that direction.
In the horizontal direction (parallel to the ramp), the horizontal force F balances the horizontal component of the gravitational force. Since the crate is moving at a constant speed, the magnitudes of these two forces must be equal.
Therefore, we can set up the following equation:
F = mg * sinθ
Substituting the given values:
F = (50 kg) * (9.8 m/s^2) * sin(31°)
F ≈ 245.9 N
So, a horizontal force of approximately 245.9 Newtons (N) is required to push the crate up the frictionless 31° ramp at a constant speed.
B) The force exerted by the ramp on the crate, also known as the normal force (N), can be determined by analyzing the vertical forces. The normal force acts perpendicular to the ramp and counteracts the downward force of gravity.
In this case, the vertical force components are the weight of the crate (mg) acting downwards and the normal force (N) acting upwards.
Since the crate is on a frictionless ramp and moving at a constant speed, we know that there is no acceleration in the vertical direction. Therefore, the magnitudes of the vertical forces must be equal.
Using trigonometry, we can determine the vertical component of the gravitational force:
Weight component perpendicular to the ramp = mg * cosθ
Since there is no net vertical force acting on the crate, the normal force (N) must balance the weight component perpendicular to the ramp.
Therefore, we can set up the following equation:
N = mg * cosθ
Substituting the given values:
N = (50 kg) * (9.8 m/s^2) * cos(31°)
N ≈ 423.6 N
So, the ramp exerts a force of approximately 423.6 Newtons (N) on the crate, in a direction perpendicular to the ramp.