Another Physics Problem!?
This question is from the forces unit:
A 50 kg crate is resting on the floor. the coefficient of friction between the floor and the crate is 0.5. a rope is attached to the crate is being pulled at an angle 25 above the horizontal. what is the largest force that can be applied to the rope that will allow the crate to stay at rest?
- MathguyLv 54 months agoFavorite Answer
I'll go first, since no one else is here...
F vertical ==> N + F sin@ - mg = 0 [ two vert forces upwards...Normal Force N and the vertical component of the rope pulling up at @ = 25˚ ... one down ]
N = normal force, F sin@ = upward component of Rope mg = weight
F = horiz forces ==> F cos@ - µN = 0 so N = ( F cos @ ) / µ
Replace N ( F cos @) / µ + F sin @ - mg = 0
you know m, µ, and @
I'll let you plug in and solve for F.... I got about 219 Newtons
- Anonymous4 months ago
Surprising how many people get their education on Yahoo! Answers. Could be why the world is so screwed up nowadays.If you can't cook, get out of the kitchen.