What Angle and Normal Force Are Involved When Pulling a Suitcase?

[jlshoop]

This may be very simple... but here is the question.

This is the question..

A woman at an airport is towing her 20.0 kg suitcase at constant speed by pulling on a strap at an angle of (theta) above the horizontal. She pulls on the strap with a 35.0 N force, and the friction force on the suitcase is 20.0 N. What angle does the strap make with the horizontal? What normal force does the ground exert on the suitcase?

I got the first part. The vertical component of the angle is 26 N. So if the suitcase is 20 kg and she is pulling vertically 26 N, shouldn't she be lifting the suitcase? how do i figure out the normal force which the ground exerts?

 

[Doc Al]

So if the suitcase is 20 kg and she is pulling vertically 26 N, shouldn't she be lifting the suitcase?

What's the weight of the suitcase? (Don't confuse 20 kg with 20 N !)

how do i figure out the normal force which the ground exerts?

Use the fact that the vertical forces on the suitcase are in equilibrium.

 

[stunner5000pt]

Ok first of all draw a free body diagram. SINCE she is moving at CONSTANT velocity, a =0 that menas the X omponents of the force are equal

then 35 cos \theta = 20

there you have theta now

Now, for the vertical components

the bag is NOT moving up or down that means the forces up and down are equal

the fac that she pulls the bag at an angle means that there is a vertical component for her force that points upward. in any case


N + 35 sin \theta - mg = 0

do you understand why N is not mg??

 

[chroot]

If the suitcase is moving at constant speed, the horizontal forces acting upon it must be equal and opposite. If there's a 20N friction force opposing its motion, the strap must be exerting 20N of force in the direction of motion.

Draw a free-body diagram. Look at the triangle made by the strap's force, and its components. Solve for theta:

\cos \theta = \frac{20}{35}

The horizontal component of the force is 20N. The vertical component of the force is not 26N.

- Warren

 

[jlshoop]

Warren- I typed the wrong thing... i meant 29 N for the vertical component, right?

another confusion i have is, is 20 kg the weight, or mass?

 

[chroot]

Yes, the vertical component is about 29N.

http://www.google.com/search?hl=en&lr=&q=35+*+sin(55.15+degrees)&btnG=Search

- Warren

 

[chroot]

another confusion i have is, is 20 kg the weight, or mass?

The kilogram is a unit of mass. To obtain the weight, you must multiply the mass by the gravitational acceleration, 9.81 m/s^2, in accordance with F=ma.

- Warren

 

[jlshoop]

so the weight is 216

N = 216 - 29 = 187

BY GEORGE THINK I'VE GOT IT...

yeah?

 

[jlshoop]

Thanks very very very much Warren... you really truly rock.

 

[chroot]

So do stunner500 and Doc Al. :)

- Warren

 

[jlshoop]

I agree with that... didn't mean to sell anybody short on their physics talent :^)