A block of mass 57.1 kg rests on a slope having an angle of elevation of 28.3°. If pushing downhill on the block with a force just exceeding 162 N and parallel to the slope is sufficient to cause the block to start moving, find the coefficient of static friction.
HINT - Draw the block's free-body diagram and apply Newton's second law for the tilted x-component, solving for μs after substituting
fs, max = μsn.
weight component down slope = m g sin 28.3
weight component normal to surface = m g cos 28.3
so friction force up slope = μ m g cos 28.3
so
162 + m g sin 28.3 = μ m g cos 28.3
To find the coefficient of static friction, we need to analyze the forces acting on the block and apply Newton's second law.
1. Draw the free-body diagram of the block on the slope. Identify the forces acting on it.
- The weight force (mg) acts vertically downward.
- The normal force (N) acts perpendicular to the slope.
- The force of static friction (fs) acts parallel to the slope and prevents the block from sliding.
2. Apply Newton's second law in the x-direction (parallel to the slope):
Fnet_x = max
The net force in the x-direction consists of the force pushing downhill (162 N) and the force of static friction (fs):
Fnet_x = fs - mg*sin(theta)
Since the block is at rest, the net force in the x-direction is zero:
0 = fs - mg*sin(theta)
3. Now we need to express the force of static friction (fs) in terms of the normal force (N) and the coefficient of static friction (μs).
fs = μs*N
Substitute this expression into the previous equation:
0 = μs*N - mg*sin(theta)
4. Rearrange the equation to solve for μs:
μs = (mg*sin(theta))/N
We need to find the value of N. By analyzing the forces in the y-direction (perpendicular to the slope), we can determine that:
N = mg*cos(theta)
Substitute this value into the equation for μs:
μs = (mg*sin(theta))/(mg*cos(theta))
Simplify the expression:
μs = tan(theta)
5. Substitute the given angle of elevation (theta = 28.3°) into the equation:
μs = tan(28.3°)
6. Use a calculator or lookup table to find the tangent of 28.3°:
μs ≈ 0.531
Therefore, the coefficient of static friction (μs) is approximately 0.531.
To find the coefficient of static friction, we need to analyze the forces acting on the block and apply Newton's second law. Let's proceed with the solution step by step:
Step 1: Draw the free-body diagram:
On the inclined plane, we have the following forces acting on the block:
- The weight of the block acting straight downward (mg).
- The normal force (N) perpendicular to the incline.
- The force of static friction (fs) parallel to the incline.
Step 2: Resolve the forces:
Resolving the weight (mg) into components parallel and perpendicular to the incline, we get:
- The component of weight parallel to the incline (mg*sinθ), pointing downhill.
- The component of weight perpendicular to the incline (mg*cosθ), pointing into the plane.
Note: θ is the angle of elevation, which in this case is 28.3°.
Step 3: Apply Newton's second law:
For the block to be on the verge of moving, the force of static friction must be at its maximum value. Therefore, the force applied downhill (162 N) is equal to the maximum static friction force (fs, max). This can be written as fs, max = 162 N.
Using Newton's second law for the x-component along the incline, we get:
fs, max = μs * N
Since N = mg*cosθ, we can substitute this into the equation:
fs, max = μs * mg*cosθ
Step 4: Substitute values and solve for μs:
Substituting the known values, we have:
162 N = μs * 57.1 kg * 9.8 m/s^2 * cos(28.3°)
Simplifying:
162 N = μs * 556.66 kg*m/s^2 * 0.882
Dividing both sides by 556.66 kg*m/s^2 * 0.882:
μs = 162 N / (556.66 kg*m/s^2 * 0.882)
Calculating the value:
μs ≈ 0.329
Therefore, the coefficient of static friction is approximately 0.329.