EK 1001 Physics Question #312

slam1924

Hey there!

Was going through a couple of EK 1001 Physics questions and came across this question that I couldn't quite figure out so hopefully someone here could lend some much needed clarification.

Here's a picture of the diagram and the question prompt: My attempt at a solution:
a. Torque (CW) = Torque (CCW) --> 0.2mg = 0.2f where m = mass of the sign; I picked the rotation point to be at the attachment point of the rope to the sign which is 0.2m away from the wall and 0.2m away from the center of the sign; mg = f
b. Since the sign is at equilibrium, forces in the y direction cancel out so Tcos30 = mg + f --> T = (mg +f)/cos30
c. Since frictional force, f, <= uN where N = normal force and u = static coefficient;
d. Solving for f from (b), we get f = Tcos30 - mg, so Tcos30 - mg <= uN
e. Solving for T, and knowing that mg = f from (a), we get T = 2mg/cos30 from (b)
f. Plugging f and T into (d), we get mg < uN

This is where I get stuck. The book claims that since forces in the horizontal direction cancel out, N = Tsin30. Shouldn't the normal force always point in the vertical direction? From my understanding of the diagram, shouldn't N = Tcos30?

Hopefully someone can clarify this for me.

2+ Year Member
Is the answer C?

OP
S

slam1924

Yep it is! After a lot of head banging, I realized that since the sign makes contact with the wall's surface in the "horizontal" direction in this case, it makes sense that N = Tsin30. Thanks anyways!

• bobeanie95