# Acceleration produced from dragging a box up a hill?

• FREEDOGGY
In summary, the conversation discusses finding the acceleration of a box being pulled across a smooth horizontal floor with a force of 20.0N at a 60 degree angle to the horizontal. The solution involves using trigonometry to find the X component of the force and then using that in the equation for acceleration.

#### FREEDOGGY

A box 80.0N is to be pulled across a smooth horizontal floor with a force of 20.0N at an angle of 60 degrees to the horizontal. What acceleration will be produced?

I know this probably seems simple, but I have searched my lecture notes, my textbook and the web but I can't find out how to input the 60 degrees into finding the acceleration. Just wondering if anybody has any hints? thanks in advance

The force has two components, one in the vertical direction and one in the horizontal direction. The vertical component will be canceled by its weight. So there will only be an acceleration in the horizontal direction. If you draw a picture of the force vector and its x-y components how do you express the x component in terms of its hypothenuse by using simple trig?

FREEDOGGY said:
A box 80.0N is to be pulled across a smooth horizontal floor with a force of 20.0N at an angle of 60 degrees to the horizontal. What acceleration will be produced?

I know this probably seems simple, but I have searched my lecture notes, my textbook and the web but I can't find out how to input the 60 degrees into finding the acceleration. Just wondering if anybody has any hints? thanks in advance
Welcome to PF.
hint:
The force has components ( X and Y).
And as You can see there is no acceleration in the Y direction ,so acceleration=F(x)/M
when F(x) component in the X direction of the force(FIDN IT USING The ANGLE!);
good luck

EDIT:Cyosis beat me XD
I'm starting to get old :<