How Much Force is Needed to Move a Box with a Rope at an Angle?

[missrikku]

Okay, a box of mass m is dragged across a floor by pulling a rope attached to the box and inclined at an angle @ above the horizontal.

given that the coefficient of static friction is 0.05, what minimum force mag is required from the rope to start the crate moving?

When it says it is inclined at an angle @, does that mean that the rope has an angle @ above the horizontal? If this is the case, could you let me know if this process is correct:

Fx = Tcos@ - fs = Tcos@ - 0.5N = ma = 0
Fy = N + Tsin@ - mg = ma = 0 ---> N = mg - Tsin@

from Fx:
Tcos@ - 0.5(mg - Tsin@) = 0

Then I can solve for T, right?

 

[HallsofIvy]

Yes, the "effective weight" is the actual weight minus the upward lift from the rope: mg- T sin θ. The force you must apply to overcome static friction is (0.5)(mg- T sin θ) and the horizontal force to do that is T cos θ : the equation is
0.5(mg- T sin θ)= T cos θ as you have.

 

[M98Ranger]

Yes, your process is correct. To clarify, when it says the rope is inclined at an angle @ above the horizontal, it means that the rope is at an angle @ with respect to the floor. So, your equations for Fx and Fy are correct. To solve for T, you can rearrange the Fx equation to get T = 0.5mg/cos@, which gives you the minimum force required from the rope to start the crate moving.