# Physics helps needed on a few

1. Nov 29, 2007

### Cowtipper

1. A woman is towing a 20.0 kg suitcase at a constant speed by pulling on a strap that is at angle Z above the horizontal. She pulls on the strap with a 35.0 newton force, and the friction force on the suitcase is 20.0 newtons. What normal force does the ground exert on the suitcase?

I have found that the angle is 55.2 degrees, and I know that the answer to the question is 167 newtons. I just don't know how to get there...

2. An object with mass m1 = 5.00 kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2 = 10.0 kg. Find the acceleration of each object and the tension in the cable.

3. m1 = 10 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30.

(a) If the system is released from rest, what will its acceleration be?

(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?

4. A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between box and floor is 0.570, how long does it take to move the box 4.00 m, starting from rest?

5. A car is traveling at 50.o km/hr on a flat highway.

(a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

(b) What ist he stopping distance when the surface is dry and the coefficient of friction is 0.600?

Thanks so much for any help. I'm not looking for answers, just how to get started and where to go...

Thanks again.

2. Nov 29, 2007

### sephirothrr

Draw free-body diagrams! Label all the forces.

1. What is the definition of the normal force? What do you need in order to find it?

For the rest of these, you really must draw free-body diagrams. They let you see what's really going on.

3. Nov 29, 2007

### Cowtipper

This is for number 2. Is this the kind of thing you're looking for?

4. Nov 29, 2007

### sephirothrr

Not quite. Draw each block separately, and draw all the forces that act on the block as arrows in their respective directions.

5. Dec 4, 2007

### Cowtipper

Okay, I have solved number two: a) 6.54 m/s^2 and b) 32.7 N.

Number three, I've solved a), it is 0. But b) still confuses me.

I've gotten this far:

T=tension
Ff=force of friction
m=mass
g=gravity
a=acceleration

T-Ff = μmg = ma
T-Ff = 0.3(10)(9.81) = 10a
T-29.43 = 10a

And that's just about as far as I've gotten on that one. Any assistance would be appreciated, thanks!

6. Dec 4, 2007

### Cowtipper

Number four has been solved.

This is number four:

4. A box of books weighing 300 N is shoved across the floor of an apartment by a force of 400N exerted downward at an angle of 35.2 degrees below the horizontal. If the coefficient of kinetic friction between box and floor is 0.570, how long does it take to move the box 4.00 m, starting from rest?

This is what I did:

FNormalVertical = 400Nsin35.2 + 300N = 530.6N
FNormalHorizontal = 400cos35.2 Ffriction = 326.9N - Ffriction
Ffriction (also known as μFNormal) = 0.570(530.6N) = 302.4N
Fnet (also known as mass*acceleration) = 326.9N - 302.4N = 24.5N
mass = 300N/9.81 = 30.6kg
24.5N = 30.6kg(acceleration) = 24.5N/30.6kg = .801 = acceleration
d=velocityinitial(t) + 1/2(acceleration)(time^2)
4meters=0(t) + 1/2(0.801m/s/s)(time^2)
4meters=1/2(0.801m/s/s)(time^2)
4meters=.4005(time^2) = 4meters/.4005 = time^2 = 9.988 = t^2 = sqrt(9.988) = t = 3.16 seconds

Woo!

These ones are still puzzling me though.

3. m1 = 10 kg and m2 = 4.0 kg. The coefficient of static friction between m1 and the horizontal surface is 0.50, and the coefficient of kinetic friction is 0.30.

(b) If the system is set in motion with m2 moving downward, what will be the acceleration of the system?

5. A car is traveling at 50.o km/hr on a flat highway.

(a) If the coefficient of friction between road and tires on a rainy day is 0.100, what is the minimum distance in which the car will stop?

(b) What ist he stopping distance when the surface is dry and the coefficient of friction is 0.600?

Thanks again.

7. Oct 20, 2009

### Loehreralec

I am having trouble with Question 5, i too was assigned this problem and after a bit of tinkering and looking at the formulas i still cannot figure out how you could get the distance it would take to stop the car with only the speed and the co-efficient of Kinetic Friction.
I figure:

fK=(.1)*N

Seeing as we are not given the Weight of the car I don't see how i could figure out the normal force...

Were i to find the normal force and then the friction that would get me the acceleration and i could just plug my variables into:

deltaX= V0*T+(.5)(A)(T2) but sadly i can't get to this step.

I'm sure im missing something blatantly obvious but thats why i need help >.< thanks