# Homework Help: Couple questions from a newbie

1. Sep 23, 2007

### rickportega

1. The problem statement, all variables and given/known data
In the drawing, the rope and the pulleys are massless, and there is no friction. (m1 = 8.4 kg, and m2 = 2.95 kg.)
(a) Find the tension in the rope.
(b) Find the acceleration of the 8.4-kg block. (Hint: The larger mass moves twice as far as the smaller mass.)

2. Relevant equations
F=ma

3. The attempt at a solution

Im not familiar with how pulleys affect masses. I guessed that (m2*9.8) was the tension in the rope, which was wrong. Then I tried (m2*9.8)/2, figuring that the end attached to the ceiling takes 1/2 the force away. Is this reasoning wrong? It seems to make sense, but it was wrong too. For the acceleration, I tried (m2*9.8)/8.4, using the formula f=ma, but I think the f was wrong.

Question #2

1. The problem statement, all variables and given/known data
A 53.6 kg sign is suspended by two wires, as the drawing shows. Find the tension in wire 1 and in wire 2.

2. Relevant equations
F=ma
sin(x)=o/h
cos(x)=a/h

3. The attempt at a solution

I broke the two wire tensions into the components, saying that the two y components add up to equal (53.6*9.8), but I think that is where I went wrong. I then solved for the other sides of the "triangle" using (x/cos(55))+(x/cos(43)) = (53.6*9.8)
Even though the y components are the same length on paper, I guess the tensions are different. Im not really sure what else to try.

More details on attempt:
I solved for the components of each. What I got was for wire 1, the y component = T1cos(43) and the x component T1sin(43)
For wire 2, the y component = T2cos(55) and the x component = T2cos(55)
From this, I wrote two equations. T1+T2 = (9.8*53.6) = 525.28 and -(T1Sin(43))+(T2(Sin(55)) = 0, figuring that the addition of the x components should equal 0, since the wires come together at a point.
Using these 2 equations, I solved for T1 = 286.636 and T2 = 238.64

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Last edited: Sep 23, 2007
2. Sep 23, 2007

### learningphysics

For question 1... use the freebody diagram of m1... get the F=ma equation in the horizontal direction. then use the freebody diagram of m2.... get the F=ma equation in the vertical direction...

The accelerations of the two masses are the same... hence if you label a as the accleeration of m1 to the right, a should also be labelled as the acceleration of m2 downwards.

For question 2... what is x? Let T1 and T2 be the tension in the two wires wires... What are the vertical components of the two tensions... what are the horizontal components?

3. Sep 23, 2007

### rickportega

the part where im having trouble is drawing the free-body for m2. The only force acting on it is (9.8*2.95) = 28.91N in the downward direction, right? But I'm not sure how the rope attached to the ceiling affects it's fall.

As for #2. I solved for the components of each. What I got was for wire 1, the y component = T1cos(43) and the x component T1sin(43)
For wire 2, the y component = T2cos(55) and the x component = T2cos(55)
From this, I wrote two equations. T1+T2 = (9.8*53.6) = 525.28 and -(T1Sin(43))+(T2(Sin(55)) = 0, figuring that the addition of the x components should equal 0, since the wires come together at a point.
Using these 2 equations, I solved for T1 = 286.636 and T2 = 238.64, however these were wrong.

Last edited: Sep 23, 2007
4. Sep 23, 2007

### learningphysics

m2 has gravity acting downwards... but there is 2T acting upward (where T is the tension in the rope)

write out the equations for m1 and m2...

You're mixing up the x and y components... T1cos(43) is the x component. Also, in the y-direction you wrote "T1 + T2 = (9.8*53.6)" why didn't you add the y-components instead of T1 and T2?

5. Sep 23, 2007

### rickportega

Sorry, this was a typo " x component = T2cos(55)" it should be
"x component = T2sin(55)"

Is that what you meant when you said I got the x and y components mixed up? I dont see anything wrong with the T1 components....sin(A)=opp/hyp, cos(A)=adj/hyp, thats right, right?

Anyway, I changed the t1+t2=525.28 equation to what you suggested and got t1=434.5 and t2=361.76, but these are both wrong still

I see what you mean about a 2T force acting upwards on m2, I didn't realize that. However, I still don't know how to find T :(

6. Sep 23, 2007

### learningphysics

The x - component of T1 is T1cos43. The y component is T1sin43... The x-component of T2 is T2cos55. The y-component of T2 is T2sin55. The equations should be:

-T1cos43 + T2cos55 = 0

T1sin43 + T2sin55-mg = 0

Please write out the equations... otherwise it is difficult for us to know what you're understanding, and missing...

7. Sep 23, 2007

### rickportega

Im not sure what the equations are. I thought it'd be f=ma, so 2.95*9.8=28.91N, however this is not the answer. Im not sure what other equation to try

8. Sep 23, 2007

### learningphysics

$$\Sigma{F} = ma$$

the sum of all the forces acting on an object is the mass of the object times the acceleration...

So apply this to m2...

What forces are summed?

9. Sep 23, 2007

### rickportega

Would it be the 9.8 from gravity, plus 2x that same force, from the upward pull of the ropes? so -(2.95*9.8)+(2(2.95*9.8))?

10. Sep 23, 2007

### learningphysics

I'm confused by that...

It would be:

2T - m2*g = m(-a), because we are taking a to be downward. You need the forces on the left side... the mass * acceleration on the right side.

What is the equation for m1?

11. Sep 24, 2007

### rickportega

well because it is a frictionless surface, m1 only has 1 relevant force acting on it, and that is of the rope. The force on it would be equal to the tension, so a = f/m.

Ugh, im so confused by this. with the equation 2T - m2*g = m(-a), don't the m2*g and m(-a) cancel out, leaving just 2T = 0?

12. Sep 24, 2007

### learningphysics

No, they don't cancel out... I highly recommend going over this section in your text... m2*g is only one force... g is not the acceleration of m2... m2*g is just the gravitational force. Tension is another force... the sum of all the forces equals ma or in this case m(-a)...

We're adding all the forces (vector sum) acting on m2 together... that result = mass*acceleration.

13. Sep 24, 2007

### physicolic

sorry but in the first question the accelerations of the masses are not the same. There is a relation between them. For example if m1 makes a displacement x, m2 will move a fraction of x at the same time interval. So the accelerations will not be the same.

14. Sep 24, 2007

### learningphysics

Ah... you're absolutely right. The acceleration of m2 is (1/2) acceleration of m1.