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These three questions are on ym homework and I don't know how to finish them.

by warfreak131
Tags: finish, homework
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warfreak131
#1
Sep27-09, 03:59 PM
P: 181
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Question 1
1. The problem statement, all variables and given/known data

1. The masses mA and mB slide on the smooth (frictionless) inclines fixed as shown in the first figure. Determine (a), a formula for the acceleration of the system in terms of mA, mB, θA, θB, and g. (b), if θA = 32, θB = 23, and mA = 5 kg, what value of mB would keep the system at rest? What would be the tension in the cord (neg. mass) in this case?


2. Relevant equations

(mB*Sin(θB)) - (mA*Sin(θA))g / (mA + mB)

3. The attempt at a solution

I had no idea how to get the equation, so I looked in the back of the book, and it says

(mB*Sin(θB)) - (mA*Sin(θA))g / (mA + mB)

How do you get that!?

I figured out how to get the mass of B, i set the equation to 0 and solved for it, that part was easy.

As for the tension, I still wasnt sure, so i checked the book, and it said 26 N. I played around with the numbers a little bit, and found out that it is the top portion of that equation divided by 2. Is that just coincidence, or it that the right way to do it? If not, what is?

Question 2
1. The problem statement, all variables and given/known data

2. A 3.0kg block sits on top of a 5.0kg block which is on a horizontal surface. The 5 kg block is pulled to the right with a force F as shown in figure 2. The coefficient of static friction between all surfaces is .6, and the kinetic coefficient is .4. (a), what is the minimum value of F needed to move the two blocks? (b), if the force is 10% greater than your answer for (a), what is the acceleration of each block?

2. Relevant equations

Ff = μ*Fn

3. The attempt at a solution

I found the force of friction needed to pull the 3kg block along the 5kg block, which is μ*Fn = (.6 * (3g)) = 17.7N

The force needed to pull the 5 kg block along the ground would be the same as above, except you use 8kg, because its the weight of both blocks combined, right?, (.6 * 8g) = 47.1 N

I figured that you would need to pull with the combined static forces of the 3 kg block and the 8kg of the two masses, which would equal 64.8 N. But the book says 82 N. And I am not sure how to get the acceleration, which the book says is 4.5 m/s

Question 3
1. The problem statement, all variables and given/known data

3. A bicyclist can coast down a 7 deg. hill at a steady 9.5km/h (approx. 2.6 m/s). If the drag force is proportional to the square of the speed v, so that Fd = -cv, calculate (a), the value of the constant c, and (b) the average force that must be applied in order to descend the hill at 25km/h (approx 7m/s). The mass of the cyclist plus the bicycle is 80kg. Ignore other types of friction.

2. Relevant equations

Fd = -cv, F = mgSin(theta)

3. The attempt at a solution

I had no idea where to start. I looked up drag force in the book, and its so terribly vague that I cant understand what theyre saying. The answers from the back of the book are c = 14kg/m, and 570N

please help me with these!
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tiny-tim
#2
Sep28-09, 03:58 PM
Sci Advisor
HW Helper
Thanks
tiny-tim's Avatar
P: 26,157
Hi warfreak131! Welcome to PF!

(not a good idea to put three long questions in the same thread! )
Quote Quote by warfreak131 View Post
1. The masses mA and mB slide on the smooth (frictionless) inclines fixed as shown in the first figure. Determine (a), a formula for the acceleration of the system in terms of mA, mB, θA, θB, and g. (b), if θA = 32, θB = 23, and mA = 5 kg, what value of mB would keep the system at rest? What would be the tension in the cord (neg. mass) in this case?


2. Relevant equations

(mB*Sin(θB)) - (mA*Sin(θA))g / (mA + mB)

3. The attempt at a solution

I had no idea how to get the equation, so I looked in the back of the book, and it says

(mB*Sin(θB)) - (mA*Sin(θA))g / (mA + mB)

How do you get that!?

I figured out how to get the mass of B, i set the equation to 0 and solved for it, that part was easy.

As for the tension, I still wasnt sure, so i checked the book, and it said 26 N. I played around with the numbers a little bit, and found out that it is the top portion of that equation divided by 2. Is that just coincidence, or it that the right way to do it? If not, what is?
Call the tension T (it will be the same on each side of the string), and the acceleration a

then do good ol' Newton's second law twice, once for each block

that gives you two equations, so eliminate T from them, to get a.
warfreak131
#3
Sep28-09, 04:40 PM
P: 181
cool, thanks

any idea about the other two?


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