- #1
Rijad Hadzic
- 321
- 20
Homework Statement
A filled treasure chest (Mass = 375 kg) with a long rope tied along its center lies in the middle of a room. Dirk wishes to drag the chest, but there is friction between the chest and the floor, with the coeffecient of static friction = .52
If the angle between the rope and the floor is 30 degrees, what is the magnitude of the tension required just to get the chest moving?
Homework Equations
[itex]f=ma[/itex]
[itex] F_static = MkFn [/itex]
The Attempt at a Solution
First I would like for everyone to see this diagram to see if it is correct:
http://i.imgur.com/mCpPKIR.png
So is it true, that the Force of static friction is going to be opposite to the tension force? Meaning the force of friction isn't going to be pointed in the -j hat direction? Because the rope is being lifted at an angle of 30 degrees, does that mean that the force of static friction has i hat and j hat components?
Working this problem out... I did:
[itex] F_g = 375 kg * 9.81 = 3678.75 N [/itex]
[itex] F_n = F_g [/itex]
[itex] F_static = F_n * .52 = 1912.95 N [/itex]
so finding the x component of [itex] F_static [/itex] :
[itex] 1912.95cos(30) = 1655.840572 [/itex]
and now y component of static :
[itex] 1912.95sin(30) = 956.475 [/itex]
Okay now this is where I'm stumped...
My books says the answer is 1700 N, but the force of static friction is 1912.95 N! wouldn't that mean that the tension has to be at least this much just to get the thing moving?