A 2.00 kg block is held in equilibrium on an incline of angle θ = 75° by a horizontal force applied in the direction shown in the figure below. If the coefficient of static friction between block and incline is μs = 0.300, determine the following.
(a) the minimum value of
N
(b) the normal force exerted by the incline on the block
N
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To determine the minimum value of N, we can start by analyzing the forces acting on the block.
Considering the forces in the y-direction (perpendicular to the incline), we can break down the force due to gravity (mg) into components:
mg * cos(θ) = N
Therefore, N = mg * cos(θ), where m is the mass of the block and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Now, let's move on to finding the normal force (N') exerted by the incline on the block. The normal force is the force exerted by a surface perpendicular to that surface. In this case, it is the force exerted by the incline on the block.
Considering the forces in the x-direction (parallel to the incline), we can analyze the forces acting on the block. These forces include the horizontal applied force (F_applied) and the force due to the friction between the block and the incline (F_friction). The force causing the block to remain in equilibrium is the force of static friction.
F_applied = F_friction
The force of friction can be calculated as:
F_friction = μs * N'
Given that the coefficient of static friction (μs) is 0.300, we can substitute it into the equation:
F_friction = 0.300 * N'
And since F_applied = F_friction, we have:
F_applied = 0.300 * N'
Now, let's solve for N'. Rearranging the equation, we get:
N' = F_applied / 0.300
It's important to note that for the block to remain in equilibrium, the applied force needs to be balanced by static friction so that there is no net force. Therefore, the applied force should be equal to or less than the maximum force of static friction.
So, to find the minimum value of N (a), we substitute the known values into N = mg * cos(θ) and find:
N = (2.00 kg)(9.8 m/s^2) * cos(75°)
Calculating this expression will give us the minimum value of N.
Similarly, to find the normal force exerted by the incline on the block (b), we substitute the known values into N' = F_applied / 0.300 and calculate the expression.
To find the minimum value of N and the normal force exerted by the incline on the block, we need to analyze the forces acting on the block.
The force diagram for the block on the incline is as follows:
|–––––––––––––––––––––|
|\ /|
| \ / |
| \ / |
| \ / |
| \ / |
| \/ |
|_______________|
Here, the vertical direction is split into two components: one parallel to the incline (denoted by N) and one perpendicular to the incline (denoted by mg * cosθ).
The forces acting on the block are:
1. Weight (mg): mg * sinθ (component parallel to the incline)
2. Normal force (N): component perpendicular to the incline
3. Applied horizontal force (F)
(a) We need to find the minimum value of N. To do this, we can use the fact that the block is in equilibrium, meaning the net force in both the horizontal and vertical directions is zero.
In the vertical direction, it can be represented as:
N - mg * cosθ = 0
Rearranging the equation, we get:
N = mg * cosθ
Substituting the values, we have:
N = (2.00 kg) * (9.8 m/s²) * cos(75°)
N ≈ 37.87 N
Therefore, the minimum value of N is approximately 37.87 N.
(b) To find the normal force exerted by the incline on the block, we can use the fact that the block is still in equilibrium, so the net force in the horizontal direction is zero.
The horizontal force acting on the block is the applied force (F). The static friction force (fs) opposes the applied force and prevents the block from sliding down the incline.
The equation for static friction force is:
fs = μs * N
Here, μs is the coefficient of static friction between the block and the incline, and N is the normal force.
Since the block is in equilibrium, the static friction force must be equal to the horizontal applied force:
fs = F
Thus, we can write:
μs * N = F
Substituting the known values, we have:
(0.300) * N = F
To find F, we need to use the trigonometric relationship between the applied force and the weight of the block. We can express F in terms of the weight (mg) and the angle θ:
F = mg * sinθ
Substituting the values, we have:
(0.300) * N = (2.00 kg) * (9.8 m/s²) * sin(75°)
Simplifying the equation, we get:
N ≈ 21.89 N
Therefore, the normal force exerted by the incline on the block is approximately 21.89 N.