What is the Angle and Normal Force in a Towing Scenario?

[bolivartech]

Homework Statement

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?

Homework Equations

F - fk = ma

W = mg

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!

 

[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?

 

[bolivartech]

y = 35N sinθ
x = 25N cosθ - 20N

?

 

[rock.freak667]

y = 35N sinθ
x = 35cosθ

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.

 

[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?

 

[rock.freak667]

ok well you should remember that the force she is pulling with has a vertical component of 35sinθ which points upwards just like the normal reaction (R).

The sum of these two = weight.

I think you can find R now.

 

[bolivartech]

Got it, thanks!