A 1520-N crate is to be held in place on a ramp that rises at 30.0° above the horizontal. The massless rope attached to the crate makes a 22.0° angle above the surface of the ramp. The coefficients of friction between the crate and the surface of the ramp are ìk= 0.450 and" ìs= 0.650. The pulley has no appreciable mass or friction. What is the maximum weight w needed to hold this crate stationary on the ramp?
1380N
Well, let's break this down, shall we?
First of all, we need to find the forces acting on the crate. We have the weight of the crate acting downwards, which is given by the mass of the crate multiplied by the acceleration due to gravity. In this case, the weight is 1520 N.
Now, let's consider the forces along the ramp. We have the normal force, which is perpendicular to the ramp, and the frictional force, which opposes the motion of the crate.
The normal force can be found by taking the component of the weight perpendicular to the ramp. In this case, it would be the weight multiplied by the cosine of the angle of inclination of the ramp, which is 30°.
The frictional force can be found by taking the component of the weight parallel to the ramp. In this case, it would be the weight multiplied by the sine of the angle of inclination of the ramp, which is 30°.
Now, we need to determine the maximum frictional force that can be exerted on the crate before it starts to move. This is given by the coefficient of static friction multiplied by the normal force.
So, the maximum weight needed to hold the crate stationary on the ramp is the sum of the normal force and the maximum frictional force.
Wait, what's that noise? Oh, it's the sound of me trying to remember my high school physics lessons. Ah well, I'll leave the math to you. Good luck!
To determine the maximum weight needed to hold the crate stationary on the ramp, we need to consider the forces acting on the crate.
Step 1: Resolve the weight of the crate into components.
The weight of the crate (1520 N) can be split into two components: one parallel to the ramp (W⊥) and one perpendicular to the ramp (W‖).
W⊥ = mgcosθ = 1520 N * cos(30°)
W‖ = mgsinθ = 1520 N * sin(30°)
Step 2: Determine the frictional forces.
The coefficient of kinetic friction (μk) is provided as 0.450. The frictional force opposing motion on the ramp (Fk) can be calculated as:
Fk = μk * W⊥
Step 3: Calculate the tension in the rope.
The tension in the rope (T) can be determined by considering the forces acting vertically on the crate:
T - W‖ - Fk = 0
Step 4: Calculate the maximum frictional force.
The maximum static frictional force (Fs) can be calculated as:
Fs = μs * W⊥
Step 5: Calculate the minimum tension needed to overcome static friction.
If the crate is at its maximum static friction limit, the minimum tension needed to overcome static friction can be calculated as:
Tmin = Fs + W‖
Step 6: Determine the maximum weight needed.
Since we want to find the maximum weight needed to hold the crate stationary, the tension in the rope should be equal to the minimum tension needed to overcome static friction.
Therefore, we can set Tmin equal to T:
Fs + W‖ = T
Solving these equations will give us the maximum weight needed (w) to hold the crate stationary on the ramp.
Note: Make sure to convert angles to radians if your calculator or formulas require it.
To find the maximum weight (w) needed to hold the crate stationary on the ramp, we need to analyze the forces acting on the crate.
Let's break down the forces into components:
1. The weight of the crate (1520 N) can be split into two components:
- The component perpendicular to the ramp: W⊥ = W * cos(30°)
- The component parallel to the ramp: W∥ = W * sin(30°)
2. The tension force in the rope can also be split into two components:
- The component perpendicular to the ramp: T⊥ = T * cos(22°)
- The component parallel to the ramp: T∥ = T * sin(22°)
Now, let's analyze the forces acting on the crate:
1. The perpendicular component of the weight (W⊥) creates a normal force (N) that opposes it. The normal force is perpendicular to the ramp.
2. The parallel component of the weight (W∥) creates a force of friction (f) that opposes it. This force acts parallel to the ramp.
3. The perpendicular component of the tension force (T⊥) also creates a normal force (N) that opposes it.
4. The parallel component of the tension force (T∥) helps in balancing the parallel component of the weight (W∥).
Since we want to find the maximum weight (w) needed to hold the crate stationary, the maximum weight occurs when the force of friction is at its maximum (fmax). The maximum force of friction (fmax) is given by the equation:
fmax = μs * N
where μs is the coefficient of static friction and N is the normal force.
To find the normal force (N), we need to consider the vertical equilibrium of forces:
N + W⊥ + T⊥ = 0
Solving this equation will give us the value of N.
Once we have N, we can substitute it into the equation for fmax and solve for the maximum weight (w):
fmax = μs * N = w * sin(30°)
Now, substitute the given values into the equations and solve for w:
1. W = 1520 N
2. μk = 0.450
3. μs = 0.650
4. Angle between the ramp and the horizontal = 30°
5. Angle between the rope and the surface of the ramp = 22°
Using these values, we can calculate the maximum weight (w) needed to hold the crate stationary on the ramp.