A 100 kg crate is pushed at constant speed up the frictionless 30 degree ramp by a horizontal force vector F. What are the magnitudes of a) vector F and b) the force on the crate from the ramp?
I am not sure how to find part a. What would I need to do to find it? for part b would one of the forces be a normal force, and others (which I am not sure of). How would I find the normal force and others.
(a) The component of the horizontal force F in the direction UP the ramp must be equal and opposite to the component of the weight along the same direction.
F cos 30 = M g sin 30
F = M g tan 30
(b) The ramp exerts both a vertical and horizontal force on the crate. The force vector must be perpendicular to the ramp since ther is no friction.
Let F' be the force you want.
Its vertical component, F' sin 30 must equal the weight, Mg.
Solve for F'
222
.6 N
To find the magnitude of the horizontal force vector F (part a), you can use the equation:
F = M * g * tan(30)
where M is the mass of the crate (100 kg), g is the acceleration due to gravity (approximately 9.8 m/s^2), and tan(30) is the tangent of the angle of the ramp (30 degrees).
Substituting the values into the equation:
F = 100 kg * 9.8 m/s^2 * tan(30 degrees)
Calculating this expression will give you the magnitude of the horizontal force vector F.
For part b, the force on the crate from the ramp consists of both a normal force and a force along the ramp. Since the ramp is frictionless, there is no horizontal force component along the ramp.
The normal force, denoted by N, cancels out the vertical component of the weight and prevents the crate from sinking into the ramp. In this case, the vertical component of the weight is M * g * sin(30). Therefore, the normal force N is equal to M * g * sin(30).
Next, we need to find the horizontal force exerted by the ramp, which is the force you're looking for. Let's call it F'.
The vertical component of the force exerted by the ramp, F' * sin(30), must equal the weight of the crate, which is M * g. Therefore, we have the equation:
F' * sin(30) = M * g
Solve this equation for F' by dividing both sides by sin(30):
F' = (M * g) / sin(30)
Substitute the known values into the equation:
F' = (100 kg * 9.8 m/s^2) / sin(30 degrees)
Calculating this expression will give you the magnitude of the force on the crate from the ramp (part b).
To find the magnitude of vector F (part a), we can use the equation:
F cos 30 = M g sin 30
where F is the magnitude of vector F, M is the mass of the crate (100 kg), g is the acceleration due to gravity (9.8 m/s^2), and 30 degrees is the angle of the ramp.
Simplifying the equation:
F * (sqrt(3)/2) = 100 * 9.8 * (1/2)
F * sqrt(3) = 490
F = 490 / sqrt(3)
F ≈ 282.84 Newtons
So the magnitude of vector F is approximately 282.84 Newtons.
To find the force on the crate from the ramp (part b), we need to calculate the normal force and the force exerted parallel to the ramp by the ramp on the crate.
The normal force is the perpendicular force exerted by the ramp on the crate. It can be calculated using the equation:
F' sin 30 = M g
Where F' is the magnitude of the force exerted by the ramp on the crate.
Simplifying the equation:
F' * 1/2 = 100 * 9.8
F' = 100 * 9.8 * 2
F' = 1960 Newtons
So the magnitude of the force exerted by the ramp on the crate is 1960 Newtons.
The other force exerted by the ramp on the crate is parallel to the ramp. Since the ramp is frictionless, this force is equal to the component of the weight of the crate parallel to the ramp. This force can be calculated using the equation:
F' cos 30 = M g cos 30
Simplifying the equation:
F' * sqrt(3)/2 = 100 * 9.8 * sqrt(3)/2
F' = 100 * 9.8 * sqrt(3)
F' ≈ 17018.67 Newtons
So the magnitude of the force exerted by the ramp parallel to the ramp is approximately 17018.67 Newtons.
To summarize:
a) The magnitude of vector F is approximately 282.84 Newtons.
b) The force on the crate from the ramp consists of a normal force of 1960 Newtons and a force parallel to the ramp of approximately 17018.67 Newtons.