an object of mass 5kg in plane on a plane inclined at an angle of 30 degree to the horizontal. calculate the force on the object perpendicular to the plane when the object is at rest.
Why is it that the horizontal is 5gcos30 not 5gsin30
5g cos30°
think about it. when the plane is horizontal, the angle is zero, and the whole weight is down against the plane (cos 0° = 1)
To calculate the force on the object perpendicular to the plane when it is at rest, we need to consider the forces acting on the object.
In this case, we have an object with a mass of 5 kg located on an inclined plane at an angle of 30 degrees to the horizontal.
The force on the object perpendicular to the plane is known as the normal force, denoted as N. This force acts in a direction perpendicular to the plane and helps support the weight of the object.
To find the normal force, we can first calculate the weight of the object. The weight is given by the formula:
Weight = mass * gravitational acceleration
Given that the mass of the object is 5 kg and the gravitational acceleration is approximately 9.8 m/s^2, we can calculate the weight:
Weight = 5 kg * 9.8 m/s^2 = 49 N
Now, the weight of the object can be resolved into two components: one parallel to the inclined plane and the other perpendicular to the plane.
The component of the weight parallel to the inclined plane is given by:
Weight_parallel = Weight * sin(θ)
where θ is the angle of inclination (30 degrees). So:
Weight_parallel = 49 N * sin(30 degrees) ≈ 24.5 N
The component of the weight perpendicular to the inclined plane is equal to the normal force:
Normal Force = Weight * cos(θ)
Normal Force = 49 N * cos(30 degrees) ≈ 42.43 N
Therefore, when the object is at rest, the force on the object perpendicular to the plane is approximately 42.43 Newtons.