a 2.0 kg crate is incline at 60degrees and is held in equilibrium by a horizontal force..
what is the magnitude of this horizontal force? (disregard friction)
The inclines applies a force M g cos 60 = M g/2 to the crate, in a direction normal to the incline surface. The horizontal component of that force equals the force that keeps it in equilibrium.
I will leave you to compute that
is the horizontal force the same as the normal force?
No. The normal force is perpendicular to the incline surface
To find the magnitude of the horizontal force required to hold the 2.0 kg crate in equilibrium on an incline at 60 degrees, we need to analyze the forces acting on the crate. Since friction is disregarded, the only forces acting on the crate are its weight and the normal force exerted by the incline.
The weight of the crate can be calculated using the equation:
Weight = mass x gravitational acceleration
Weight = 2.0 kg x 9.8 m/s^2 (gravitational acceleration)
Weight = 19.6 N
Now, let's break down the weight force into its components. The force can be split into two perpendicular directions - one parallel to the incline (the horizontal direction) and one perpendicular to the incline (the vertical direction).
The component of the weight force perpendicular to the incline is given by:
Weight perpendicular = Weight x cos(angle of incline)
Weight perpendicular = 19.6 N x cos(60 degrees)
Weight perpendicular = 9.8 N
The component of the weight force parallel to the incline is given by:
Weight parallel = Weight x sin(angle of incline)
Weight parallel = 19.6 N x sin(60 degrees)
Weight parallel ≈ 16.98 N
Since the crate is held in equilibrium, the horizontal force must counterbalance the component of the weight force that is parallel to the incline. Therefore, the magnitude of the horizontal force is approximately 16.98 N.