Find Angle to Pull Toaster Cord for Min. Tension

[pf21avs]

Homework Statement

A 1.13 kg toaster is not plugged in. The coefficient of static friction between the toaster and a horizontal countertop is 0.395. To make the toaster start moving, you carelessly pull on its electric cord. For the cord tension to be as small as possible, you should pull at what angle above the horizontal?

Homework Equations

FBD maybe a sum of all forces problem

The Attempt at a Solution

in my fbd i have come to the point where in the x direction i have Ff=Ftcos(theta) and in the y direction i have fn=fg-ftsin(theta), well i know i can substitute the fn equation for Ff=(mu)fn however once I'm done simplifying i get ftcos(theta)=(mu)Fg-(mu)Ftsin(theta). two unknowns and one equation, i went to the physics tutors office and he told me that fn disappears but when i asked why he just said its in the question and left me to myself, i feel as though i have expended all resources and was wondering if someone could help with this question pleasE?

note: i apologize for the lack of grammerical correctness

 

[Arbitrary]

A free body diagram is a good start. However, I believe this problem is solvable without all those equations--think about it for a moment. When is the tension force the smallest and the toaster still moving?

Consider what happens when the tension force is at an angle of 90 degrees to the horizontal. The component of the tension force that makes it move would be Ft*cos (90 degrees). Since cos (90 degrees) = 0, then no part of the tension force is making the object move, so the object is not going to move. Then consider what would happen if the angle of the tension force to the horizontal is 90...

Good luck. ;)

 

[Vijay Bhatnagar]

You have written the equation correctly. Next apply the condition for minimum T with respect to theta (differential of T with respect to theta = 0). Differentiate the equation you have already written wrt theta and in that put dT/dtheta = 0. Solve for theta. [Try. If still in doubt post again with the work done].