M1 - Friction - Help! I really can't do M1

Can someone clarify this for me: What does it actually mean to "resolve" a force???

6) A straight footpath makes an angle of x-degrees with the horizontal. An object P of weight 1250 N rests on the footpath. The coefficient of friction between the object and the footpath is 0.1. The least magnitude of a force, acting upon the footpath, which will ho0ld the object at rest on the footpath is 50 N. By treating the object as a particle, show that the value of x satisfies 10sinx - cosx = 0.4

9) A small toy has mass 1.2 kg. When a child tries to move the toy along a horizontal floor by pushing with a force of magnitude 5N, acting downwards at an angle of 30 degrees to the horizontal, she is unsuccessful. When the child tries to move the toy by pulling with a force of magnitude 5N, acting upwards, she is successful. Show that the coefficient between the toy and the floor lifes between 0.30 and 0.47 (2sf)

I manged to get:

5cos30 = 14.26(mu)
mu = 0.30

but i'm not really sure why it's that..

Please could someone help me - and explain every step in extreme detail! Thanks

Reply 1

Chewwy

17

force along slope due to weight = 1250sinx

friction(max) = coeff. of rest. * force perp. to slope = 0.1 * 1250cosx (in the other direction to the weight, so can throw a minus on the front of it)

we're interested about when the resultant force is equal to 50N, i.e. the object is on the point of slipping...

1250sinx - 125cosx = 50

(divide by 125)

10sinx - cosx = 0.4

Reply 2

Kawada89

i see you managerd to get mu= 0.30
i cant get that, i'm confused about the 'When the child tries to move the toy by pulling with a force of magnitude 5N, acting upwards, she is successful'

is that 5N straight up? if so does friction matter...

but anyways, i managed to get the 0.47...

taking the first push by the child, to no prevail:

this is saying the friction is greater than the push:

horizontal push<Friction
5cos30<(mu)x reaction

reaction is found by resolving the forces vertically, (by putting all up equal to all down)

so R + 5sin30 = 1.2g
R = 1.2g - 5sin30
R = 11.76 - 2.5
R = 9.26

baack to our original equation:
5cos30<(mu)x9.26
(mu)>5cos30/9.26
giving
(mu)>0.47 (2sf)
now although this is greater than which, and we want less than, it is the number we're looking for..

so either there is a minus in the inequality, and so the sign is reversed, or im jsut wrong.. hopefully combining this with how you got (mu) = 0.30 will help you

maybe not... :smile:

Reply 3

nota bene

15

Kawada89

baack to our original equation:
5cos30<(mu)x9.26
(mu)>5cos30/9.26
giving
(mu)>0.47 (2sf)
now although this is greater than which, and we want less than, it is the number we're looking for..

so either there is a minus in the inequality, and so the sign is reversed, or im jsut wrong.. hopefully combining this with how you got (mu) = 0.30 will help you

maybe not... :smile:

I have not checked through the question, but if 5cos30<9.26mu then
5cos30/9.26>mu

edit: erase nonsense...

Reply 4

Kawada89

nota bene

I have not checked through the question, but if 5cos30<9.26mu then
5cos30/9.26>mu

but 5cos30<9.26mu yes
then 5cos30/9.26<mu as we bring the 9.26 under?

i then turned it around to get mu on the LHS, hence switching the inequality:
mu>5cos30/9.26
mu>0.47

but im not too sure, seeing as im already in the wrong :s-smilie: