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)
3 answers
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 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.
b. The coefficient of friction is approximately 0.577.