Posted by Andrew on Thursday, January 17, 2013 at 7:26pm.
Start with a picture if you don't already have one and then draw a free body diagram (a dot with arrows to represent all the forces applied to the box if you don't know what that is go ahead and google it- it'll make sense if you see a picture). We can assume up and the direction of the applied force (F) is positive.
Forces in the y direction:
+Fn (normal force)
-mg (mass*gravity)
-F*sin(theta)
Fsin(theta) is the componet of the applied force in the y direction. If you remember the trig identites sin(theta)=opposite/hypotenuse. Here, the applied force is the hypo. so you get Fsin(theta)= the y component of the applied force since it is ppposite the angle at which the force is applied to the box.
X direction:
+Fcos(theta)
This is the same as the y component of the applied force except its the adjacent side to the angle and acts in the x direction.
Ok if that all made sense (I hope it did) now you're ready to write two equations, one in the x direction and one in the y, based on Newton's second law F=ma
Y direction:
Fn-mg-Fsin(theta)=ma
X direction:
Fcos(theta)=ma
Since you have all the information in the x direction substitute with numbers and solve for a. It should come out to be something around 2.47 m/s^2 if you rounded the same way I did.
PS if there was friction you would subtract its force from the x direction when using sum of forces = ma.
Hope this helps!