A body of weight 6N rest on a plane inclined at 30 the original calculate
a. the Force that keep it from sliding down the plane
b. what coefficient of friction (N)
To calculate the force that keeps the body from sliding down the plane, we first need to find the component of the weight force that acts parallel to the inclined plane. This can be done using trigonometry.
The weight force can be split into two components: one perpendicular to the inclined plane (mg * cos(theta)) and one parallel to the inclined plane (mg * sin(theta)), where m is the mass of the body and g is the acceleration due to gravity.
Given that the weight of the body is 6N, we can calculate the mass using the equation F = mg, where F is the weight:
6N = m * 9.8 m/s^2
m = 6N / 9.8 m/s^2
m ≈ 0.612 kg
Now, we can find the component of the weight force parallel to the inclined plane:
Force parallel = mg * sin(theta)
Force parallel = 0.612 kg * 9.8 m/s^2 * sin(30°)
Force parallel ≈ 3.02 N
a. The force that keeps the body from sliding down the plane is approximately 3.02 N.
To find the coefficient of friction, we need to use the force of friction formula:
Force of friction = coefficient of friction * Normal force
The normal force is the force perpendicular to the inclined plane, which can be found using trigonometry:
Normal force = mg * cos(theta)
Normal force = 0.612 kg * 9.8 m/s^2 * cos(30°)
Normal force ≈ 5.33 N
Substituting the values into the formula, we can solve for the coefficient of friction:
3.02 N = coefficient of friction * 5.33 N
coefficient of friction ≈ 0.57
b. The coefficient of friction is approximately 0.57.
A body of weight 6N rest on a plane inclined at 30 the original calculate
a. the Force that keep it from sliding down the plane
b. what coefficient of friction (N)
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component of weight down slope = 6 sin 30 = 3 Newtons so we need 3 Newtons up slope
component of weight against plane below = 6 cos 30 = 5.2 Newtons
so
mu * 5.2 = 3
mu = 3/5.2 = 0.577
a. The force that keeps the body from sliding down the plane is approximately 3 Newtons.
b. The coefficient of friction is approximately 0.577.
To calculate the force that keeps the body from sliding down the inclined plane, we need to consider the forces acting on the body.
a. The forces acting on the body are:
- The weight of the body acting vertically downwards (W = 6N).
- The normal force (N) acting perpendicular to the inclined plane.
Since the plane is inclined at an angle of 30°, we can break down the weight of the body into two components:
- The component acting parallel to the plane (W //), given by W // = W * sin(30°)
- The component acting perpendicular to the plane (W ⊥), given by W ⊥ = W * cos(30°)
The force that keeps the body from sliding down the inclined plane is equal to the component of the weight acting parallel to the plane. Therefore:
- Force to keep it from sliding down = W // = W * sin(30°) = 6N * sin(30°)
b. To calculate the coefficient of friction (μ), we need the normal force (N) acting perpendicular to the plane. The normal force is equal to the component of the weight acting perpendicular to the plane. Therefore:
- N = W ⊥ = W * cos(30°) = 6N * cos(30°)
The coefficient of friction (μ) is the ratio of the frictional force (F) to the normal force (N). In this case, the frictional force is the force required to keep the body from sliding down the plane. Hence:
- μ = F / N = (6N * sin(30°)) / (6N * cos(30°))
Simplifying further:
- μ = sin(30°) / cos(30°)
Using trigonometric identities, we know that:
- sin(30°) = 0.5
- cos(30°) = (√3) / 2
Substituting these values in the equation:
- μ = (0.5) / ((√3) / 2)
- μ = 0.5 * (2 / (√3))
- μ = 1 / (√3)
Hence, the coefficient of friction (μ) in this case is 1 / (√3).