Kinetic Fricion Acceleration Problem

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laxxie
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Homework Statement


To move a large crate across a rough floor, you push on it with a force at an angle of 21 degrees, find the acceleration of the crate if the applied force is 400 , the mass of the crate is 32 and the coefficient of kinetic friction is 0.43.

A = ?
Force = 400N
Theta = 21 Degrees
Mass = 32Kg
Kin fricion coefficient = .43



Homework Equations


Force of Kinetic Friction = Kinetic Friction Coefficient x Normal Force
and probably some random 2 dimension kinematics equation to solve for acceleration...


The Attempt at a Solution


I basically know that if you are pushing down on the box at an angle of 21 degrees than you are basically adding to the weight of hte box and therefore increasing the normal force acting on the box so I did cos21 * 400 + 32kg*9.81 to get the normal force and we already have the static fricion coefficient so then I just multiplied those together to get the kinetic friction force which is wrong so I'm not sure what mistake I'm making here. If you guys could help me out that would be great my teacher is awful and I need to learn this.
 
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EDIT: nvm; I figured it out, but PhanthomJay can have this one.
 
Last edited:
If the angle of the applied force is 21 degrees with the horizontal, then it's 400*sin21 that must be added to the weight. If the angle of the applied force is 21 degrees with the vertical, then your calculation for the kinetic friction force, acting opposite the motion , is correct. But then you have to find the acceleration using Newton's 2nd law, where the sum of forces in x direction (that is, the net force in the x direction) = ma, where the net force is always in the direction of the acceleration.
 
Thanks, was using wrong trig function haha