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## Homework Statement

All variables are given in the pictures.

## Homework Equations

Nothing more than simple force equations.

## The Attempt at a Solution

I attempted it on my own, but I don't know if the answers are right.

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- Thread starter dwangus
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- #1

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All variables are given in the pictures.

Nothing more than simple force equations.

I attempted it on my own, but I don't know if the answers are right.

- #2

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- #3

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- #4

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I see no indication that you have made any attempt at solving these. If you want help, show what you have done and indicate where you are stuck.

- #5

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^^^ I don't know if I'm stuck, that's why I need to know whether or not my answers are wrong.

I self-teach myself a lot, let's just put it that way.

- #6

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Show your answers and how you got them. You seem to misunderstand this forum. People here will go out of their way to help you learn how to solve problems, but you are asking to be simply spoon-fed answers and that doesn't fly here.

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Ok, but...

In order to complete physics hw, being spoon-fed answers just doesn't cut it. Amongst all subjects, physics requires work shown AND an answer to even remotely do well in class.

So if you're concerned that I'll just take your answers and run with them and put them on my worksheet and call them my own, it seems a bit redundant because I wouldn't even get a good grade on the homework.

So I don't think "spoon-fed" is the correct term here; I don't think cynical behavior can exist when just asking for answers to a physics worksheet.

- #8

Doc Al

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It's quite simple:

Show

We don't just hand out answers. (Read our rules, which are linked at the top of every page.)

In general, don't pile a bunch of problems into the same thread (or post!); take them one at a time.

- #9

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Tension in second rope between 3kg and 4kg abbreviated as = T2

40 - T2 = 4a

T2 - 30 = 3a

a = 10/7

T2 = 240/7

T = T2 + mg

T = 240/7 + 40 = 520/7

- #10

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For the 6 kg mass,

T + N (exerted from the 3kg mass) = 60N

For the 3 kg mass,

T = 30 + N (exerted from the 6kg mass)

2T = 90

T (also from the pin) = 45N

Normal Force = 15N

- #11

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Problem 3 (Previous Problems were 1 and 2):

Acceleration (A) = -Acceleration (B)

T - m(A)g = 5a(A)

T - m(B)g = 4a(B)

T - 50 = 5a(A) -----> 50 - T = -5a(A) = 5a(B)

T - 40 = 4a(B)

---------------------------

10 = 9a(B)

Acceleration of B = 10/9

Acceleration of A = -10/9

- #12

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Problem 4:

FBD of mass m =

-Force of mg downwards

-Force of μ(s)F(app) upwards to cancel out mg

-Force applied to the right

-Opposite normal force (N1) exerted by mass M to the left

μ(s)F(app) = 50

F(app) = 125

F(app) - N1 = 5a(m)

N1 = 10a(M)

F(app) = 15a

Acceleration of pair = 8.33333

Normal Force of pair = 150N

- #13

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Problem 5:

Gravity Force exerted from 2kg mass and 20kg mass = 220N

Total Normal Force = Gravity Force + Net Force of 8kg mass = 220 + 40 = 260N

I feel as though my reasoning is far too simple to have gotten this one right.

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Problem 6:

Acceleration of mass A = Acceleration of mass B (downwards is considered negative in the case of mass B for convenience)

20 - T = 2a(B)

T - μ(k)m(A)g = 2a(A)

T - 4 = 2a(A)

-----------------

16 = 4a

a(A and B) = 4m/s^s

T = 12N

(mass of C and A)g + [(mass of B)(g) - T] = Normal Force of system

70 + 20 - 12 = 78

(Normal Force system)μ(s) = (Normal Force of A on C)μ(k)

78μ(s) = 4

μ(s) = 0.5128

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Problem 7: (a-c: force is not counted in problem yet)

a) Since it has velocity up the incline, the frictional force is added (helping) to mgsin30

30cos(30)(0.3) + 30sin(30) = 3a

a = 7.598

b) Velocity is down the incline, so frictional force is subtracted (opposing) from mgsin30

a = 2.402

c) Since there is no movement, frictional STATIC force is subtracted from mgsin30

a = 1.534

d) The resultant force that opposes mgsin30 in this case to keep the mass in equilibrium is F/cos30

Therefore, the actual force applied F = mgsin30cos30 = 12.99N

e) The minimum force to hold the block in equilibrium is WITH the help of friction to slow down the mgsin30, so

mgsin30 = F/cos30 + (0.4)30cos30

F = 3.99N

Maximum force is with friction OPPOSING to slow down Force applied, so

mgsin30 + 0.4(30cos30) = F/cos30

F = 21.99N

- #16

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People here generally go out of their way to be helpful but you are making this difficult. I would suggest that you ask a moderator to delete this thread and then you start a new thread for each of your problems, showing all of your work for the problem in one place with the statement of the problem. the way you have it now, you are asking people to go to a bit of trouble just to figure out what you've done so far. You might get lucky and someone will go to the trouble, but you'd likely have more luck if you separate the problems.

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....how am I making this difficult?

- #18

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I was just trying to be helpful. If you don't get it you don't get it. Just hope for the best

- #19

Doc Al

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Since you've included 7 problems in a single thread, an extended discussion of each will get jumbled. Much easier if you put them in separate threads. Or at least did one at a time.....how am I making this difficult?

- #20

Doc Al

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Problem 1:

This is good.Tension in second rope between 3kg and 4kg abbreviated as = T2

40 - T2 = 4a

T2 - 30 = 3a

a = 10/7

T2 = 240/7

Two rope segments pull down on the upper 4kg mass.T = T2 + mg

- #21

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Like I said, can anyone just tell me if I'm right or wrong?

It's ok if you don't tell me how I'm wrong, I want it that way.

This way, I can chance upon the answer myself.

- #22

Doc Al

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- #23

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^^^ welp then, looks like a bunch of my other problems are wrong then

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