I am studying physics and need help with inclined planes. How do you know if you use sin or cos?
Also could you please make a mock example question if you have time? thank you :)
Also could you please make a mock example question if you have time? thank you :)
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When you have a mass, m kg, resting on an inclined plane, with a coefficient of static friction, μ, and an angle of inclination Ѳ degrees, the technique is to analyze the forces involved. All of the forces come from the weight of the mass. So
weight = mg, and this ALWAYS acts VERTICALLY DOWNWARDS.
Since weight is a force, it can be resolved into its components. For ease of calculation, the normal method is to resolve it in a direction at right angles to the plane, and parallel to the plane.
Component of weight at right angles to plane = mg cosѲ
Component of weight parallel to the plane = mg sinѲ
It will help if you draw a simple sketch. Start with the plane horizontal. Imagine that there is a weight on the end of a piece of string that is attached to a nail that sticks out at right angles to
the underside of the plane. When the plane is horizontal, the string and nail line up. As you tip the plane, can you see that the angle between the string and the nail will be the same as the angle of the plane? From that you can see how we get the components of the weight of a mass resting on the inclined plane.
By definition, the frictional force, which ALWAYS opposes any motion, is given by
F(r) = μ R
where μ = coefficient of static friction; R = reaction force of the plane on the mass.
The reaction force is ALWAYS taken at right angles to the plane. So
R = m g cosѲ
Also
μ = tan Ѳ
Example Question
The coefficient of static friction of hard rubber on a street surface is μ = 0.8. What is the steepest slope on which an automobile with rubber tyres ( and locked wheels} can rest without sliding?
Ans
Since the car is not sliding, the frictional force acting up the slope must be just balancing the component of the weight of the car acting down the slope. So we can say
μ mg cosѲ = mg sinѲ
μ = sinѲ / cosѲ = tanѲ
Ѳ = arc tan (0.8) = 38.66 degrees
weight = mg, and this ALWAYS acts VERTICALLY DOWNWARDS.
Since weight is a force, it can be resolved into its components. For ease of calculation, the normal method is to resolve it in a direction at right angles to the plane, and parallel to the plane.
Component of weight at right angles to plane = mg cosѲ
Component of weight parallel to the plane = mg sinѲ
It will help if you draw a simple sketch. Start with the plane horizontal. Imagine that there is a weight on the end of a piece of string that is attached to a nail that sticks out at right angles to
the underside of the plane. When the plane is horizontal, the string and nail line up. As you tip the plane, can you see that the angle between the string and the nail will be the same as the angle of the plane? From that you can see how we get the components of the weight of a mass resting on the inclined plane.
By definition, the frictional force, which ALWAYS opposes any motion, is given by
F(r) = μ R
where μ = coefficient of static friction; R = reaction force of the plane on the mass.
The reaction force is ALWAYS taken at right angles to the plane. So
R = m g cosѲ
Also
μ = tan Ѳ
Example Question
The coefficient of static friction of hard rubber on a street surface is μ = 0.8. What is the steepest slope on which an automobile with rubber tyres ( and locked wheels} can rest without sliding?
Ans
Since the car is not sliding, the frictional force acting up the slope must be just balancing the component of the weight of the car acting down the slope. So we can say
μ mg cosѲ = mg sinѲ
μ = sinѲ / cosѲ = tanѲ
Ѳ = arc tan (0.8) = 38.66 degrees