Pulling object at an angle problem

[xasuma]

Homework Statement

A woman at an airport is dragging her 9.0 kg suitcase at a constant speed by pulling a strapa at an angle of θ above the horizontal. She pulls the streap with a 24.0 N force, and the friction force on the suitcase is 7.5 N.

a)Draw a free body diagram of the suitcase
b)What angle does the strap make with the horizontal?
c)What normal force does the ground exert on the suitcase?

Homework Equations

Fa=cosθ-fk=0

The Attempt at a Solution

a) This is what I got. Should 7.5 N (friction) point directly to the left instead? If so, why? http://puu.sh/chxIH/b42d9f8b68.png

b) I think this might be it:
cosθ-fk=0
cosθ = 7.5N/24N
θ = cos^-1 (7.5/24)
θ = 71.79`

c) This is what I am trying:

n = mg sinθ
n= 9.0kg*9.8m/s*sin71.79`
n= 83.78NPlease let me know if something is wrong/how to fix it. Thank you

 

[NascentOxygen]

Friction acts along the rough surface. So friction is always perpendicular to N.

You then ignored your incorrect diagram, and got the right answer in (a). Though your first line in (a) is not right.

You are nowhere near being right in (b). Look at it again, once you draw the right diagram.

 

[xasuma]

Ok so for a this is it then:

a)http://puu.sh/chQHU/7023dd03b8.png

note: on the post above did you mean 'b' and 'c' , instead of 'a' and 'b' respectively? .
if so:
b)θ = 71.79`

and

c) would normal force = mg , since there is no movement in the vertical axis?
if so: 9.0kg*9.8m/s = 88.2 N ?

 

[haruspex]

note: on the post above did you mean 'b' and 'c' , instead of 'a' and 'b' respectively? .

I feel sure that's what NascentO meant.

c) would normal force = mg , since there is no movement in the vertical axis?
if so: 9.0kg*9.8m/s = 88.2 N ?

What forces acting on the case have a vertical component? What is the vertical component of each?

 

[NascentOxygen]

c) would normal force = mg , since there is no movement in the vertical axis?
if so: 9.0kg*9.8m/s = 88.2 N ?

Weight acts vertically, so is normal to the surface, yes. But the rope conveys a vertical component of force, too, in addition to its horizontal component.

 

[xasuma]

Ok so I am thinking I use Pythagorean theorem to combine them

therefore:

sqrt(((mgcos71.79)^2)+((mgsin71.79)^2))
= 88.20 N

I just realized that is the same as mg.
I am confused on this one, to me it makes sense that the normal force = mg, since there is no vertical movement. But I feel something is wrong.

 

[NascentOxygen]

There is a vertical force acting downwards on the body, at the same time there is a force delivering an upwards action to the body. These will partly cancel.

 

[xasuma]

then:

vertical component + normal force = mg?
mgsin71.79 + n = 88.2N
n = 4.42N

 

[NascentOxygen]

That doesn't mean anything to me.

You should draw the free body diagram showing forces in equilibrium. One force is weight, acting downwards. Another is friction. Etc. Once you have all the forces sketched, their vectors must all add to zero, showing equilibrium.

 

[haruspex]

to me it makes sense that the normal force = mg, since there is no vertical movement. But I feel something is wrong.

The normal force and mg are only two of the three forces acting on the case which have a vertical component. It is the combination of all three that produces equilibrium.

 

[xasuma]

I know that. In my previous post I attempted to combine the 3 of them (normal force, the 'y' component of the external force, and mg).
I understand you are trying to make me figure it out on my own but this is a bit frustrating. Thank you all for your help but I'll call this one off.

 

[haruspex]

then:

vertical component + normal force = mg?

Yes.

mgsin71.79 + n = 88.2N

mg sin θ for the vertical component of the strap tension? How do you arrive at that?