Refrigerator magnitude Question

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To calculate the magnitude of the horizontal force F applied to a refrigerator being pushed up a ramp at constant speed, the forces acting on the refrigerator must be analyzed. The weight of the refrigerator can be decomposed into components along the ramp, with F_x equal to mg*sin(θ) and F_y equal to mg*cos(θ). The applied force F can be resolved into its component parallel to the ramp, represented as F' = F*cos(θ). For the refrigerator to move at constant speed, F' must equal the gravitational component along the ramp, leading to the relationship F*cos(θ) = mg*sin(θ). Ultimately, this results in the conclusion that the magnitude of the horizontal force F is mg*tan(θ).
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(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
 
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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?
 
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