# Determine angle of force

1. Sep 22, 2009

### bolivartech

1. The problem statement, all variables and given/known data

A woman at an airport is towing her 20.0-kg suitcase at constant speed by pulling on a strap at an angle θ above the horizontal (Fig. P5.40). She pulls on the strap with a 35.0-N force. The friction force on the suitcase is 20.0 N. Draw a free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?

2. Relevant equations

F - fk = ma

W = mg

3. The attempt at a solution

(a) F - fk = ma

35.0 N – 20.0 N = (20.0 kg)(sinѲ)(9.8 m/s2)

sinѲ = (15.0 N) / (20.0 kg)(9.8 m/s2)

Ѳ = sin-1[(15.0 N) / (20.0 kg)(9.8 m/s2)]

Ѳ = °

(b) W = mg

W = (20 kg)(9.8 m/s2)

W = 196 N

I don't know if what I attempted was even on the right path. Could someone point me in the right direction? Thanks!

2. Sep 22, 2009

### rock.freak667

the 35N at an angle of θ, has what x and y components?

It says it pulls at constant velocity which means the resultant horizontal force is what?

3. Sep 22, 2009

### bolivartech

y = 35N sinθ
x = 25N cosθ - 20N

?

4. Sep 22, 2009

### rock.freak667

So now

in the x direction the resultant force is zero so what is the equation for this sum of the forces in the x direction? You can find θ now.

5. Sep 22, 2009

### bolivartech

sorry that was a typo

so

0 = 35N cosθ - 20N

20N / 35N = cosθ

θ = cos^-1 (20N / 35N) = 48.19

I really over complicated that. I need to remember to make sure I see all the parts before I begin. Can you tell me if I did the second part correctly? The normal force is the opposite of W since it is not moving in the Y direction right?

6. Sep 22, 2009

### rock.freak667

7. Sep 22, 2009

### bolivartech

Got it, thanks!