Mechanics 1.

Hi, I'm revising for the M1 retake in November and really need help with the following questions.

Review Exercise 2
Q3, 7, 8, 11, 13 and 22.

If you could show me how to complete any of them I'd be pretty damn greatful. I've done every other question in the book, lol.

(7 and 22 are pretty identical, so showing me how to do one should be enough for me to figure out the other)

Thanks in advance.

Reply 1

Fermat

You couldn't post any of the questions, could you?

Reply 2

SunGod

3. A particle P of mass 7m is placed on a rough horizontal table, the coefficient of friction between P and the table being mu. A force of magnitude 2mg, acting upwards at an acute angle alpha to the horizontal, is applid to P and equilibrium is on the point of being broken by the particle sliding on the table. Given that tan alpha = 5/12, find the value of mu.

7. A rough plane is inclined at an angel alpha to the horizontal. where tan alpha = 3/4. A particle slides with acc. 3.5ms^-2 down a line of greatest slope of this inclined plane. Calculate the coefficient of friction between the particle and the inclined plane.

8. Same as 7 except acc. = 3ms^-2 and the plane is at 40 degrees to the horizontal. mu should be to 2 sigfig.

11. A hot-air balloon, its occupants and ballast have total mass 2000kg. The balloon is travelling horizontally with constant velocity. The occupants release 100kg. Neglecting air resistance find, in ms^-2 to 2 sigfig, the magnitude of the acceleration of the balloon.

13. A block of mass 3kg is pulled along a rough horizontal floor by a constant force of magnitude 20N inclined at an angel of 60 degrees to the UPWARD VERTICAL. The acceleration of the block has magnitude 2ms^-2. Calculate to 2 decimal places, the value of the coefficient of friction between the block and the floor.

22. Same as 7 and 8 except you are given that the particle starts from rest and covers 3.5m in 2s to work out the acceleration, the plane is inclined at 30 degrees to the horizontal this time.

Don't worry too much about showing me how to do 8 and 22. Showing me 7 should be enough to let me work out how to do 8 and 22. (Sorry if I'm asking too much)

Thanks in advance.

For number 3)

That's what the answer in the book is. I understand how you did it as well. Thanks alot :smile:

For number 7)

SunGod

11. A hot-air balloon, its occupants and ballast have total mass 2000kg. The balloon is travelling horizontally with constant velocity. The occupants release 100kg. Neglecting air resistance find, in ms^-2 to 2 sigfig, the magnitude of the acceleration of the balloon.

F=ma
if its travelling at constant speed so the gravitational force equals the upwards force

F=gm = (9.8)(2000) = 19600N (upwards force)

if they release 100kg the mass left now is 1900kg

the downward force now is F=(9.8)(1900) = 18620N

so extra upwards force = 19600 - 18620 = 980N

F=ma
980N=1900a

a = 980/1900 = 0.52 ms^-2

Reply 7

Ben.S.

Fermat

For number 7)

God, you have such 'mathsy' writing! Did you use a quill?!?

Ben

Reply 8

SunGod

lol indeed.

I've now learnt how to do 3 and 7.

(for 7 you made a mistake, the acc was meant to be 3.5ms^-2. You probably confused it with q8. But I redid it using your method and 3.5ms^-2 and got the right answer)

I've also figured out how to do 8 and 22 thanks to your help Fermat.

Thanks to IntegralNeo for helping with 11.

I'm working on 13 right now. Hopefully I'll be able to finish it without any extra help :smile:

Thanks again guys.

Reply 9

Fermat

SunGod

lol indeed.

...

(for 7 you made a mistake, the acc was meant to be 3.5ms^-2. You probably confused it with q8. But I redid it using your method and 3.5ms^-2 and got the right answer)
....

Ah yes, I'm afraid my eyes ain't what they used to be :smile:

Reply 10

SunGod

Hmm, seems I can't do it ...

Can anyone help me out with Q13?

Reply 11

Fermat

SunGod

Hmm, seems I can't do it ...

Can anyone help me out with Q13?

Resolve the applied force (20N) into horizontal and vertical components.
The Normal Reaction is now the weight of the block less the vertical component.
The accelerating force on the block is the horizontal component less the friction.
The friction is ?? -> µ.

I get µ=0.58

Reply 12

SunGod

Your answer is right. I'm gonna do a little more work. I'm pretty sure what you're saying is what I'm doing, lol. (don't worry about posting how to do it)