A crate of mass m = 85 kg is pushed at a constant speed up a frictionless ramp (θ = 26°) by a h?
A crate of mass m = 85 kg is pushed at a constant speed up a frictionless ramp (θ = 26°) by a horizontal force . The positive direction of an x axis is up the ramp, and the positive direction of a y axis is perpendicular to the ramp. (a) What is the magnitude of F ? (b) What is the magnitude of the normal force on the crate?
I used F=85*9.8sin26=365 and got it wrong
for normal force Fn=85*9.8sin26
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The X and Y axis for this problem run parallel and perpendicular to the incline. This means that the horizontal force (F) must be broken up into it's X and Y components. The angle of this horizontal force is 26degrees below the incline. This means that X component of this force runs along the incline. The X component equals:
F x cos 26
Since the crate is moving at constant velocity, this means that the net sum of forces acting on the crate equal zero. The other force acting along the X axis is the force of gravity directed down the incline. This force is equal to:
m x g x sin26
Set these forces equal to solve for F (horizontal force)
F x cos26 = m x g x sin26
m = mass of crate
g = acceleration due to gravity
Solve for F
F = m x g x sin26 / cos26
F = 85kg x 9.80m/s^2 x sin26 / cos26 = 406N
Now let's solve for the normal force (n). On the incline the normal force is equal to a component of the gravitational force:
n= m x g x cos 26
But we also have a component of the the 406N force acting in the vertical direction. Because the angle of the 406N force is directed below the X axis, the vertical component of this force will increase the normal force. This component is equal to:
F x sin26
Add this component with the component to find the normal force on the crate.
n = F x sin26 + m x g x cos26
n = 406N x sin26 + (85 x 9.80m/s^2 x cos26) = 927N
Hope this helps.