A 50-kg block is being pulled up a 30.0^ slope by force of 250N which is parallel to the slope, but the block does not slide up the slope. What is the minimum value of the coefficient of static friction required for this to happen?
To determine the minimum value of the coefficient of static friction required for the block to not slide up the slope, we need to analyze the forces acting on the block.
Let's break down the forces into their components:
1. The force of gravity acts vertically downward with a magnitude of mg, where m is the mass of the block and g is the acceleration due to gravity. In this case, m = 50 kg and g = 9.8 m/s^2.
So, the force of gravity = 50 kg * 9.8 m/s^2 = 490 N.
2. The force pulling the block up the slope has a component parallel to the slope. This force is given as 250 N.
3. The normal force acts perpendicular to the slope. It is equal in magnitude and opposite in direction to the vertical component of the force of gravity. Therefore, the normal force is 490 N.
4. The force of static friction acts parallel to the slope and opposes the motion of the block. Its direction is upward because the block is not sliding up the slope.
Now, let's analyze the forces along the slope:
1. The force of gravity has a component parallel to the slope, which is mg * sin θ, where θ is the angle of the slope. In this case, θ = 30 degrees.
So, the parallel component of the force of gravity = 490 N * sin 30° = 245 N.
2. The force parallel to the slope is 250 N.
3. The force of static friction acts upward along the slope, opposing the motion of the block. Its magnitude is given by μs * N, where μs is the coefficient of static friction and N is the normal force.
Since the block is not sliding up the slope, the force of static friction must exactly balance the sum of the parallel components of gravity and the applied force. Therefore, the force of static friction = 245 N + 250 N = 495 N.
Now we can determine the minimum value of the coefficient of static friction (μs):
μs * N = 495 N
μs * 490 N = 495 N (since N = 490 N)
Dividing both sides by 490 N:
μs = 495 N / 490 N
Calculating the value:
μs ≈ 1.01
Therefore, the minimum value of the coefficient of static friction required for the block to not slide up the slope is approximately 1.01.
To determine the minimum value of the coefficient of static friction required for the block not to slide up the slope, we need to analyze the forces acting on the block.
Let's start by drawing a free-body diagram of the block on the incline:
|\
| \
| \
| \
| \
| mg Fapll \
| / /// \
| / /// \
| / / / \
|/ / / \
|------------------------\---------
In the diagram:
- mg represents the weight of the block (mass x acceleration due to gravity).
- Fapll represents the force parallel to the slope.
- fs represents the force of static friction.
- θ represents the angle of the slope.
From the given information, we know:
- Mass of the block (m) = 50 kg
- Force parallel to the slope (Fapll) = 250 N
- Angle of the slope (θ) = 30.0°
Now, let's analyze the forces in the horizontal (x) direction. To prevent the block from sliding up the slope, the force of static friction (fs) must counterbalance the force parallel to the slope (Fapll).
The force of static friction can be calculated using the equation:
fs = coefficient of static friction * Normal force
Since the block is on an incline, the normal force is not equal to the weight (mg), but the component of the weight perpendicular to the slope:
Normal force = mg * cos(θ)
Now, substituting the values into the equation:
fs = coefficient of static friction * mg * cos(θ)
We need to find the minimum value of the coefficient of static friction. This occurs when the force of static friction is at its maximum:
fs(max) = coefficient of static friction * Normal force(max)
The maximum frictional force occurs when the block is on the verge of sliding up the slope, and fs(max) = Fapll:
fs(max) = Fapll
Now we can set up the equation:
coefficient of static friction * mg * cos(θ) = Fapll
Substituting the given values:
coefficient of static friction * 50 kg * 9.8 m/s^2 * cos(30.0°) = 250 N
Simplifying the equation:
coefficient of static friction = 250 N / (50 kg * 9.8 m/s^2 * cos(30.0°))
Calculating the value:
coefficient of static friction ≈ 0.573
Therefore, the minimum value of the coefficient of static friction required for the block not to slide up the slope is approximately 0.573.
M*g = 50 * 9.8 = 490 N. = Wt. of block. = Normal force(Fn).
Fp = 490*sin30 = 245 N. = Force parallel with the incline.
Fs = u*Fn = u * 490 = 490u.
Fap-Fp-uFn = M*a.
250 - 245 - 490u = M*0 = 0,
u = 0.010.