Question

The masses mA and mB slide on the smooth (frictionless) inclines fixed as shown in the figure (Figure 1) .

A) Determine a formula for the acceleration of the system in terms of mA, mB ,θA ,θB, and g.

B) If θA = 34∘, θB = 25∘, and mA = 4.6kg , what value of mB would keep the system at rest?

C) What would be the tension in the cord (negligible mass) in this case?

D) What ratio, mA/mB, would allow the masses to move at constant speed along their ramps in either direction?

Answer 1

This is Figure 1.

Answer 2

Ahhhh I dunno :c Physics so hard!!
So many weird forces and stuff! Ahhhhh!

Answer 3

It's okay, thank you for looking at it anyway. =)

Answer 4

@whpalmer4

Answer 5

Have you drawn a force decomposition diagram yet?

Answer 6

I've tried to, but I am not 100% sure if it's right.

Answer 7

welcome to the club :-)

Answer 8

I know that there will be angles involved, and I just not 100% on whether it is cos or sin of the angle that I will be using.

Answer 9

Acceleration would be
sine(theta), that would be the effect of gravity pulling the block parallel to the ramp

Answer 10

Multiply that by the mass and then you've got it!

Answer 11

Okay, now wouldn't I have to subtract one of the masses from the other one? As in the acceleration in block A subtracted from block B or B from A?

Answer 12

And then divide that by the total mass of A and B?

Answer 13

Just find the force that each block is making, find the difference, divide by gravity, that's your mass

Answer 14

Ah, okay. That makes sense.

Answer 15

\[\frac{ mBsin(θB)−mAsin(θA) }{ mA+mBg } g\]

Answer 16

This is what the answer is. I had it right the first time, but the online Physics site we use didn't like the fact that I had the g in the numerator.

Answer 17

I think that's right

Answer 18

Whoops, there is not suppose to be a g in the denominator. I miss entered it into the equation this on here.

Answer 19

XD

Answer 20

I still had the correct answer on the homework though. The system is so picky. - _ -"

Answer 21

ok

Answer 22

if you want a subscript, do m_B or m_{block} (the latter form is necessary if you want more than one character in the subscript)

Answer 23

I just copied and pasted from the answer box on my homework, so the subscripts did not carry over.

Answer 24

Just remember that sin is any force or distance directed on the y-axis. The cos is any force or distance directed along the x-axis.

And also remember that the sin and cos originate from the center (0,0) and it is a unit circle and you draw a right triangle with starting point at (0,0). Like on your drawing, the left θA, then cosA = adjacent side / hypotenuse . If you know the angle, say 40 degree. then cos40 = adj / hyp. You would use cos if you wanted to know what the adjacent side was (it is the x axis) and you knew the hypotenuse length or force.

Same with the sin40 = opposite side / hypotenuse. You would use the sin of 40 degree when you knew the hypotenuse length and wanted to find the opposite side length, which is the y-axis direction.

If you didn't know the hypotenuse length or force but yo knew one of the sides then you use , tan 40 = opp /adj .
The point is , if you know the angle, you can use sin, cos or tan to find the x axis distance , the y- axis distance , and the hypotenuse distance, as long as you know the length of one of the sides in the triangle.

In problem1a , you will use sinθ because the masses going down the ramp are getting their force from gravity which is on the y-axis direction so you will use the sinθ of both masses since gravity acts on both in the y direction:

A) Determine a formula for the acceleration of the system in terms of mA, mB ,θA ,θB, and g.

Determine a formula for the acceleration of the system in terms of mA, mB ,θA ,θB, and g.
Use Newtons 2nd law for equation of motion.....

a = acceleration =[( mB *g* sinθB) - ( mA *g* sinθA)] / (mA + mB)
------------------------------------------------

B) If θA = 34∘, θB = 25∘, and mA = 4.6kg , what value of mB would keep the system at rest?
it would be if acceleration = zero:
acceleration =[( mB *g* sinθB) - ( mA *g* sinθA)] / (mA + mB)
0 = [( mB *9.81* sin25∘) - ( 4.6kg *9.81* sin34∘)] / (4.6kg + mB)
solving for mB:
mB = 6.08 kg
--------------------------

C) What would be the tension in the cord (negligible mass) in this case?

Here is the first two... going to post it because this site looked like it was going to lose my work..

Answer 25

This is what I ended up with:

Answer 26

I just checked the tension and it was wrong according to MasteringPhysics.

Answer 27

The ratio is correct though.

Answer 28

The answer was 25N because there are two chords and the 50N was the sum of the two tensions.

Answer 29

C) What would be the tension in the cord (negligible mass) in this case?
tension = T = ( 4.6kg *9.81* sin34∘) = 25.2 N

Since there is no acceleration then the tension is also equal on the other side
T = ( 6.08kg *9.81* sin25∘) = 25.2 N
---------------------------------------

Answer 30

Welp, I finished the problem. Thank you guys for all of the help! =)

Answer 31

D) What ratio, mA/mB, would allow the masses to move at constant speed along their ramps in either direction?

The current ratio where they have no velocity or acceleration is:
mA/mB = 4.6 / 6.08 = 0.756 ratio

If the ratio is greater than 0.756, then moves toward A
If the ratio is smaller than 0.756, then moves toward B

Answer 32

Ah, I always got the tension results wrong in the old days, too :-)