Solve for Pulling Force for Sled w/ Mass, Angle, & Coeff. of Friction

[nina123]

Homework Statement

A father gives his daughter a ride on a sled by applying a force F at an angle θ = 26.0° with respect to the horizontal. If the coefficient of static friction is µs = 0.153, and the combined mass of sled and child is m = 24.0 kg
What is the minimum amount of force required to move the sled from rest by
pulling it

Homework Equations

F=ma

The Attempt at a Solution

I have tried multiple ways and i keep getting the wrong answer,
I think i should take the sum of all the forces in the x direction and of the y direction and somehow get the force i need but I am not exactly sure.

HELP PLEASE?

 

[runningninja]

The words "minimum amount of force" and "from rest" imply that this is a statics problem. In statics, what is the special condition on the net force? What does the word "static" imply about the value of the acceleration in f=ma?
Once you figure that out, you should have your net x force components (which requires trig) sum to this special value for the net force as well as you net y force components (which also requires trig).
EDIT: Also make sure you have your pulling force in the correct direction. Did you draw a free body diagram?

 

[nina123]

The words "minimum amount of force" and "from rest" imply that this is a statics problem. In statics, what is the special condition on the net force? What does the word "static" imply about the value of the acceleration in f=ma?
Once you figure that out, you should have your net x force components (which requires trig) sum to this special value for the net force as well as you net y force components (which also requires trig).

So "static" would imply that the value of acceleration =0 ?right?

so i have in the x direction:
F(pull,x) + F(friction,x) +W(x) + n(x) = 0
but there is no W or F(gravity) in the x direction so that equals zero and same for n(x) (normal force) =0.
therefore i have:
F(pull,x) +f(friction,x)=0 right?

In the y direction:
F(pull,y) + F(friction,y) +W(y) + n(y) = 0
F(friction,y)=0
therefore i have:
F(pull,y) +W(y) + n(y) = 0 right?

then F(pull,x) = F(pull)cosθ
and F(pull,y) =F(pull)sinθ

I all that I've said is correct now, what would i do now? or is there other things that i am missing ?

 

[runningninja]

Since the force is pulling, is your pulling force positive or negative? Is your weight positive or negative?
I see two equations and two unknowns (Fpull and normal). Pull out your algebra toolkit and solve it.

 

[nina123]

thanks for your help, i got the answer :P