[Qmint]

Hi guys,

I'm stuck on this question (Question 4), I got part a right, but I don't understand what they did for part b? I feel it could be a mark scheme mistake but I hate that reasoning and its probably just me
Here's the paper. its question 4b)
http://www.ocr.org.uk/download/pp_11...gce_472801.pdf

All help appreciated

[Fleximetrics]

Ok so you resolve vertically to work out the reaction force:

mg-tsin(10)=r Where m is the mass of p

r = 2.615N

Resolve horizontally to find friction:

Fr= tcos(10)-ma (using f=ma)

Fr = 1.78N


Coefficient of fricition = Fr/r

= 1.78/ 2.62
= 0.68

[Qmint]

(Original post by Fleximetrics)
Ok so you resolve vertically to work out the reaction force:

mg-tsin(10)=r Where m is the mass of p

r = 2.615N

Resolve horizontally to find friction:

Fr= tcos(10)-ma (using f=ma)

Fr = 1.78N


Coefficient of fricition = Fr/r

= 1.78/ 2.62
= 0.68

Why would it be mg-tsin10. Wasn't tension going right so we you resolve vertically, the force would be downwards. So wouldn't it be 1.87cos10 + 0.3g = R

Also when you worked out Fr you ignored the 2N force
However your answer is correct according to the ms. I'm a lil' confused lol. Could you explain please

Thanks your help btw its appreciated.

[Fleximetrics]

(Original post by Qmint)
Why would it be mg-tsin10. Wasn't tension going right so we you resolve vertically, the force would be downwards. So wouldn't it be 1.87cos10 + 0.3g = R

Also when you worked out Fr you ignored the 2N force
However your answer is correct according to the ms. I'm a lil' confused lol. Could you explain please

Thanks your help btw its appreciated.

OK for this question we must look only at the 0.3kg particle. The particle is moving to the left therefore, friction is to the right and tension acts towards the left. So vertically, downwards we have mg and upowards we have the vertical component of the tension (tsin10) therefore the resultant is mg-tsin(10). If the tension acted downwards then the motion would be going to the right.

We ignore the 2N force because it doesnt act of the 0.3kg particle, as i said we isolate the particle in this question and so the 2N force is not important.

feel free to ask for more help if needed

[Fleximetrics]

(Original post by Qmint)
Why would it be mg-tsin10. Wasn't tension going right so we you resolve vertically, the force would be downwards. So wouldn't it be 1.87cos10 + 0.3g = R

Also when you worked out Fr you ignored the 2N force
However your answer is correct according to the ms. I'm a lil' confused lol. Could you explain please

Thanks your help btw its appreciated.

Its also impotant to know that tension works both ways in a string so it works to the right for the block but to the left for the particle.

[Qmint]

(Original post by Fleximetrics)
OK for this question we must look only at the 0.3kg particle. The particle is moving to the left therefore, friction is to the right and tension acts towards the left. So vertically, downwards we have mg and upowards we have the vertical component of the tension (tsin10) therefore the resultant is mg-tsin(10). If the tension acted downwards then the motion would be going to the right.

We ignore the 2N force because it doesnt act of the 0.3kg particle, as i said we isolate the particle in this question and so the 2N force is not important.

feel free to ask for more help if needed

I think I get it but one little question, just to be sure, do we treat it as isolated because it says "between P and the surface". Because if its isolated I see why tension would be left. As obi its moving left and Fr is right so T must be left.

[Fleximetrics]

(Original post by Qmint)
I think I get it but one little question, just to be sure, do we treat it as isolated because it says "between P and the surface". Because if its isolated I see why tension would be left.

yes, that is exactly why, it is possible to work out by considering both, but it requires more work than the question is worth.

I hope i have helped

Good luck

[Qmint]

(Original post by Fleximetrics)
yes, that is exactly why, it is possible to work out by considering both, but it requires more work than the question is worth.

I hope i have helped

Good luck

Yes you have thanks for you help