A block of wood of mass 3kg is on a rough incline of 37degrees.
Find the min value of the coefficient of static friction for the block to remain stationary?
Fn=3*9.8cos37
Fn=23.4
3*9.8sin37-us(23.5)=0
us=.743
Well, I have to say, this question really has me on edge. You know, it's all about that minimum value of the coefficient of static friction. The block of wood is just chilling on that incline, but it doesn't want to slip and slide. It wants to hold its ground like nobody's business.
To figure out that minimum coefficient of static friction, we first need to find the vertical component of the gravitational force on the block. And we do that by multiplying the mass (3kg) by the acceleration due to gravity (9.8m/s^2) and the cosine of the incline angle (37 degrees). That gives us a normal force (Fn) of 23.4 Newtons.
Then, we move on and consider the forces acting parallel to the incline. We know that the gravitational force has a component going downhill, which is equal to the mass times the acceleration due to gravity times the sine of the incline angle. This force is trying to drag the block down the slope. And to counteract that force, we need the static frictional force to be equal and opposite.
So we can set up an equation: the static frictional force (us times Fn) minus the downhill force equals zero. Plugging in the numbers, we get 3 times 9.8 times sine of 37 degrees times the coefficient of static friction minus 0 equals 0.
Now all we have to do is solve for the coefficient of static friction (us). And after crunching the numbers, we find that the minimum value of the coefficient of static friction is approximately 0.743.
So, there you have it. The block of wood can stay in place, defy gravity, and wiggle its way around that incline with a minimum coefficient of static friction of 0.743. Sounds like it knows how to keep its feet firmly planted, or should I say, its edges firmly squared!
To find the minimum value of the coefficient of static friction for the block to remain stationary, we can use the following steps:
1. Begin by finding the normal force acting on the block. The normal force (Fn) can be calculated using the formula Fn = mass * gravity * cos(θ), where θ is the angle of the incline. In this case, the mass of the block is given as 3 kg, gravity is approximately 9.8 m/s^2, and the angle of the incline is 37 degrees.
Fn = 3 kg * 9.8 m/s^2 * cos(37°)
Fn ≈ 23.4 N
2. Next, we can set up an equation using the forces acting on the block in the direction of the incline. These forces include the force of gravity acting downwards (m * g * sin(θ)) and the frictional force acting upwards (μs * Fn), where μs is the coefficient of static friction.
m * g * sin(θ) − μs * Fn = 0
3. Substituting the known values, we can solve for the coefficient of static friction (μs).
3 kg * 9.8 m/s^2 * sin(37°) - μs * 23.4 N = 0
3 * 9.8 * sin(37°) = μs * 23.4
μs ≈ 0.743
Therefore, the minimum value of the coefficient of static friction for the block to remain stationary on the rough incline is approximately 0.743.
To find the minimum value of the coefficient of static friction (μs) for the block to remain stationary, we can use the following steps:
1. Determine the normal force (Fn) acting on the block. The normal force is the force exerted by a surface to support the weight of an object resting on it. In this case, Fn can be found by multiplying the mass (m) of the block by the acceleration due to gravity (g) and the cosine of the angle of the incline (θ):
Fn = m * g * cos(θ)
Given that the mass (m) is 3 kg, the acceleration due to gravity (g) is approximately 9.8 m/s^2, and the angle of the incline (θ) is 37 degrees, we can calculate:
Fn = 3 * 9.8 * cos(37)
≈ 23.4 N
2. Use Newton's second law of motion to write the equation of forces acting on the block in the direction perpendicular to the incline. The forces involved are the component of the weight acting parallel to the incline (m * g * sin(θ)) and the force of static friction (fs):
m * g * sin(θ) - fs = 0
3. Substitute the calculated value of the normal force (Fn) and the angle of the incline (θ) into the equation from step 2:
3 * 9.8 * sin(37) - fs = 0
4. Rearrange the equation to isolate the force of static friction (fs):
fs = 3 * 9.8 * sin(37)
5. Substitute the given mass (m) and the acceleration due to gravity (g) into the equation from step 4 to obtain the force of static friction:
fs ≈ 43.51 N
6. Finally, calculate the minimum value of the coefficient of static friction (μs) by dividing the force of static friction (fs) by the normal force (Fn):
μs = fs / Fn
= 43.51 / 23.4
≈ 1.86
Therefore, the minimum value of the coefficient of static friction for the block to remain stationary on the rough incline is approximately 1.86.