Refrigerator magnitude Question

[courtrigrad]

(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

 

[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?

 

[quicknote]

for providing a free body diagram and setting up a coordinate system for this problem. To understand how to get the answer of mg\tan\theta , we need to use Newton's second law, which states that the net force on an object is equal to its mass times its acceleration. In this case, since the refrigerator is moving at a constant speed, its acceleration is zero. Therefore, the net force on the refrigerator must also be zero.

Using this information, we can set up an equation for the forces in the x-direction:

F_{x} = F - mg\sin\theta = 0

Solving for F, we get:

F = mg\sin\theta

This is the magnitude of the applied force in the x-direction. However, we are interested in the total magnitude of the applied force, which includes the component in the y-direction as well. So we need to find the magnitude of the total force vector, which can be calculated using the Pythagorean theorem:

|F| = \sqrt{F_{x}^2 + F_{y}^2}

Substituting in the values we know, we get:

|F| = \sqrt{(mg\sin\theta)^2 + (mg\cos\theta)^2}

Simplifying, we get:

|F| = \sqrt{m^2g^2(\sin^2\theta + \cos^2\theta)}

Using the trigonometric identity \sin^2\theta + \cos^2\theta = 1, we get:

|F| = \sqrt{m^2g^2}

|F| = mg

Therefore, the magnitude of the applied force is simply equal to the weight of the refrigerator, which is mg. This means that the answer of mg\tan\theta is incorrect. The correct answer is just mg, as we have shown above.