# Refrigerator Question

1. Nov 22, 2005

(1) A refrigerator with mass m is pushed up a ramp at constant speed by a man applying a force F . The ramp is at an angle $$\theta$$ above the horizontal. Neglect friction for the refrigerator. If the force F is horizontal , calculate its magnitude in terms of m and $$\theta$$.

Ok so I drew a free body diagram. The forces acting on it are the applied force F , the weight of the crate $$w = mg$$, and the normal force n . I set up a coordinate system in which the side parallel to the ramp is the x-axis, and the side perpendicular to the ramp is the y-axis. So I decomposed the weight vector into its following components: $$F_{x} = mg\sin\theta$$, and $$F_{y} = mg\cos\theta$$. I know the answer is $$mg\tan\theta$$. How would you get this knowing the components of the weight vector?

Thanks

2. Nov 22, 2005

### mezarashi

You have a good start. So the forces of interest are those along the parallel direction of the inclined ramp. By resolving the gravitational force mg, we get along the ramp direction $$F_g_x = mgsin\theta$$.

Now, to resolve the horizontal force F being applied to the refrigerator. If we denote a force $$F'$$ to represent the component of this force parallel to the ramp's incline, then we can see the relationship: $$Fcos\theta = F'$$. Noticeably then, for no acceleration along the ramp, then $$F' = F_g_x$$. Solved?

Last edited: Nov 22, 2005