Friction force applied at an unknown angle to horrizontal plane

[nialltm1991]

Homework Statement

im having problems with this question...a body weighing 51kg is pulled on a horrizontal plane with coefficient of friction 0.3. a force of 300N is applied at angle x ... find the angle x?

Homework Equations

The Attempt at a Solution

the only solution i hav come to is ... cosx + sinx = 0.5 but i can't solve it anymore ... can someone help me please?

 

[issacnewton]

can you set up free body diagram...let F=300N , F is applied at angle x, so there is one upward component and one forward component...also normal reaction from the ground is acting upward on body and gravitational force mg is acting downwards...set up FBD with this info and use the fact the fact that net force in both vertical and horizontal directions is zero

 

[nialltm1991]

yes i can i have a fbd ... but then i will have 2 equations with sin in one and cos in the other and that is the part i get stuck at

 

[issacnewton]

can you write your equations here so i can check them

 

[nialltm1991]

forces up = forces down forces left = forces right
Rn + 300sin(theta) = mg F = 300cos(theta)
Rn + 300sin(theta) = 500.31 muRn = 300cos(theta) because F = muRn
Rn = 300cos(theta)/mu

300cos(theta)/0.3 + 300sin(theta) = 500.31


and that's as far as i can go

 

[issacnewton]

what you can do is square both sides and then express \cos^2(x) in terms of
\sin^2(x) and then you have a quadratic equation in \sin(x)
letting y=\sin(x), solve the quadratic equation. but i did that and the values
of y i got are -1.66 and 1.935 . since sine function takes values between -1 and 1 , there is no real valued solution for the x. so please solve the quadratic equation on your own and confirm my calculations...if you are getting the same thing , then the problem is either stated incorrectly or there is some missing information.