Newton Applications. Help solving equation with sin and cos.

[rokas]

Homework Statement

At an airport, you pull a 18-kg suitcase across the floor with a strap that is at 45 degrees above the horizontal. Find the normal force and the tension on the strap, given that the suitcase moves with a constant speed and the coefficient of kinetic friction is 0.38.

[tex]\mu[/tex]k = 0.38
m = 18kg
[tex]\sum[/tex]F = 0 (because speed is constant)
angle = 45
N = ?
T = ?

Homework Equations

Fcos([tex]\theta[/tex]) - F_f = 0
F_f = mg - Fsin([tex]\theta[/tex])([tex]\mu[/tex]k)

The Attempt at a Solution

Fcos([tex]\theta[/tex]) - [mg - Fsin([tex]\theta[/tex])([tex]\mu[/tex]k)]
Fcos(42) - [(18)(9.81) - Fsin(42)](0.38) = 0

I'm trying to solve for F but I don't know how to get rid of, solve equations with cos and sin.
Help is appreciated!

 

[xeno_gear]

you just need to evaulate cos(42) and sin(42) using your calculator. Be very careful with roundoff error (don't round too much too soon), and make sure your calculator is in degrees mode (if the problem is stated in degrees.. most intro physics problems are though)

 

[rokas]

so Fcos(45) turns to .707107F?

 

[xeno_gear]

Yeah, but [tex] \cos(45^\circ) = \frac{\sqrt{2}}{2} [/tex], so maybe you can leave it like that until the very end, then you can approximate with decimals. It'll save you the hassle of round-off error. To be honest, it won't make much difference in this case, especially since it looks like you're rounding out to 6 places anyway, but it's not a bad idea in general.

but the short answer is yes.

 

[rokas]

I solved the equation,

F = 68.763N
then i found normal force
N = 127.957N

I have to find the tension on the strap, but i don't completely understand how to do that. Could someone explain the concept or how to find it?

 

[rokas]

The question I have is should I add Force and Force of Friction since you have to overcome Force of Friction?
But Force is on an angle, so does that mean i have to find Fx and then add them, and then combine Fx and Fy?

 

[HallsofIvy]

The horizontal component of force must be equal to the friction force so that there is no acceleration or deceleration.

 

[rokas]

Okay, so how do I find tension on the strap?

 

[Doc Al]

Okay, so how do I find tension on the strap?

Set up two equations, one for vertical forces and one for horizontal force. You'll have two equations and two unknowns (Normal force and Tension)--which works out just right.

 

[rokas]

I already know the normal force.
I was trying to figure it out by myself and thought if i did this:
Ty = 68.763sin(45)
Tx = 68.763cos(45) + (127.957) (0.38)
solve and combine x and y components to have the tension? not sure...at all.

 

[Doc Al]

I was trying to figure it out by myself and thought if i did this:
Ty = 68.763sin(45)

Looks to me like you already know the tension, if you're able to find it's y component. This equation says Ty = T sin(45) = 68.76 sin(45)

 

[rokas]

The 68.783n is not the tension, its the force of the pull. Or is it the same thing?
If it is the same thing, i have to add the force of friction to the x component of force since that's working against the movement?

 

[Doc Al]

The 68.783n is not the tension, its the force of the pull. Or is it the same thing?

The force with which the strap is pulled is the tension.

If it is the same thing, i have to add the force of friction to the x component of force since that's working against the movement?

The sum of the forces in the x-direction will equal zero, if that's what you mean.

 

[rokas]

Okay, thanks Doc Al!
Sorry, my teacher is pretty confusing sometimes.