[swirlwood]

im trying to get started with my m1 revision and well, having quite a bit of trouble.
http://www.teachnet-uk.org.uk/2007%2...%20answers.pdf
Q5 here has completelt confused me, im usually ok with questions like this but i cant gseem to get my ehad around this one

The way i see it theyve given us the co-efficient of friction to be 1/3, so Fr would simply be 1/3R right? and R would be the perpendicular force downwards, which is 1.5gcos30. What am i not seeing here?

Please help!

[ghostwalker]

Originally Posted by swirlwood

im trying to get started with my m1 revision and well, having quite a bit of trouble.
http://www.teachnet-uk.org.uk/2007%2...%20answers.pdf
Q5 here has completelt confused me, im usually ok with questions like this but i cant gseem to get my ehad around this one

The way i see it theyve given us the co-efficient of friction to be 1/3, so Fr would simply be 1/3R right? and R would be the perpendicular force downwards, which is 1.5gcos30. What am i not seeing here?

Please help!

The tension in the string also has a component perpendicular to the plane.

[mathslover786]

Tension has a force also

[swirlwood]

I know.Ive considered that, but I dont understand the solution. Why ahve they considered the 1/3 value for the vertical components when Fr is actually acting horizontally? =/

[ghostwalker]

Originally Posted by swirlwood

I know.Ive considered that, but I dont understand the solution. Why ahve they considered the 1/3 value for the vertical components when Fr is actually acting horizontally? =/

As far as I can see they haven't.

They resolve horizontally, which gives Fr in terms of T.

Then they resolve vertically, which gives R in terms of T.

Then they use the Fr=muR to link the two together and find T.

Does that help?