Mechanics M1: Post your doubt here

[iKhaled]

http://olevel.sourceforge.net/papers/9709/9709_s12_qp_41.pdf
Q7 plzz paper after 3 days cant figure out his question
leadingguy
iKhaled

aaaaaaahhhhhhh nice qstn still wrking.. it took more than half an hour yet. still unsolved.

gonna help u both with it in sometime i am kinda busy now..sorry!!

 

[hussamh10]

gonna help u both with it in sometime i am kinda busy now..sorry!!

Ahh no problem take your time..........

 

[leadingguy]

http://olevel.sourceforge.net/papers/9709/9709_s12_qp_41.pdf
Q7 plzz paper after 3 days cant figure out his question
leadingguy
iKhaled

hey got it...... we have to find the tension in Tc and Ta.
form 2 eq.
Tc.Cosx + Ta.Cosx = 8N
Tc.Sinx = Ta.Sinx agree??? till here

now as the lengths are given we can see that it is not an isosceles bt it is a right angled... so find the hypotnuse which can also be said as the resultant of the two tensions, Tc and Ta. refer the sketch below.

now sinx will be (2/2.5 ) as the formula implies sinx = perp./hyp. so put these values and carry on the question u will get the ans. correct.

 

[leadingguy]

the resultant is not 0.5 it is 2.5 i am sorry pleas do not confuse hussamh10

 

[hussamh10]

hey got it...... we have to find the tension in Tc and Ta.
form 2 eq.
Tc.Cosx + Ta.Cosx = 8N
Tc.Sinx = Ta.Sinx agree??? till here

now as the lengths are given we can see that it is not an isosceles bt it is a right angled... so find the hypotnuse which can also be said as the resultant of the two tensions, Tc and Ta. refer the sketch below.

now sinx will be (2/2.5 ) as the formula implies sinx = perp./hyp. so put these values and carry on the question u will get the ans. correct.
wouldnt be the equation like this
Tc sinX=Ta sinX+The weight of da ringg??

 

[hussamh10]

the resultant is not 0.5 it is 2.5 i am sorry pleas do not confuse hussamh10

wouldnt be the equation like this
Tc sinX=Ta sinX+The weight of da ringg??

 

[leadingguy]

wouldnt be the equation like this
Tc sinX=Ta sinX+The weight of da ringg??

i too took that befor but aftr lots of pondering i jxt reached to conclude that the weight of the ring has been adjusted/overcomed by the tension in the string Ta as we can say that untill the force of 8N was nt present there the ring was going down. and now it HAS BEEN BALANCED/equilibrised by the tension in the string

 

[hussamh10]

the resultant is not 0.5 it is 2.5 i am sorry pleas do not confuse hussamh10

man i got every thing just cant understand why Ta Sinx=Tc sinx isnt the weight also acting downwards

 

[hussamh10]

i too took that befor but aftr lots of pondering i jxt reached to conclude that the weight of the ring has been adjusted/overcomed by the tension in the string Ta as we can say that untill the force of 8N was nt present there the ring was going down. and now it HAS BEEN BALANCED/equilibrised by the tension in the string

could not have done that in paper surely God help us

 

[leadingguy]

could not have done that in paper surely God help us

lol practice here and GOD will make it easy there.... GOD demands efforts

 

[daviruss]

wouldnt be the equation like this
Tc sinX=Ta sinX+The weight of da ringg??

look an advice if he ever asked u to find abt both tensions u resolve as leadingguy did, but if he asked the tension for AB u resolve on the ring and thats when u take the weight i hope this helped!!!!
btw leadingguy u did the best way to solve the question (RESOLVING)!

 

[iKhaled]

man i got every thing just cant understand why Ta Sinx=Tc sinx isnt the weight also acting downwards

did u solve it already or u need help? i am free now sorry for being late

 

[hussamh10]

look an advice if he ever asked u to find abt both tensions u resolve as leadingguy did, but if he asked the tension for AB u resolve on the ring and thats when u take the weight i hope this helped!!!!
btw leadingguy u did the best way to solve the question (RESOLVING)!

yeah thats the right way thanks

 

[hussamh10]

did u solve it already or u need help? i am free now sorry for being late

can u do the question and explain whu TasinX=Tc sinX

 

[leadingguy]

question no. 3. xplain why tsinx + tcosx = 15.5... becuase 15.5 is the x component so only tsinx = 15.5N and tcosx should be = 8.5.. but there is something i am unable to get. Mr Me

 

[Mr Me]

How is your guys preparation going on ?

 

[AhShun]

question no. 3. xplain why tsinx + tcosx = 15.5... becuase 15.5 is the x component so only tsinx = 15.5N and tcosx should be = 8.5.. but there is something i am unable to get. Mr Me

Here it is

How is your guys preparation going on ?

It's going fine

 

[daviruss]

Here it is


It's going fine

if u knw how to resolve then ur on the safe side haha resolving for those of question is the best

 

[iKhaled]

can u do the question and explain whu TasinX=Tc sinX

7 (i)
you have a right angle triangle with height 2m and base of 1.5 m and 2 unknowns angle. make one angle called theta and another angle called alpha as i showed u in the triangle i showed u

now lets to some trigs,

cos theta = 2/2.5
sin theta = 1.5/2.5

cos alpha = 1.5/2.5
sin alpha = 2/2.5

i resolved Ta and Tc and notice that the angle that Ta makes with the vertical is equal to alpha and the angle that Tc makes with the vertical is equal to theta. now its time to build up some equations!

Tc cos theta = Ta cos alpha
Tc (2/2.5) = Ta (1.5/2.5)
0.8 Tc = 0.6 Ta

Tc sin theta + Ta sin alpha = 8
Tc ( 1.5/2.5) + Ta (2/2.5) = 8
0.6 Tc + 0.8 Ta = 8

Ta = 0.8Tc/0.6

0.6Tc + 0.8(0.8Tc/0.6) = 8...solve this equation and u will get Tc = 4.8N

Ta = 0.8(4.8)/0.6
Ta = 6.4 N

(ii) i guess that's easy to solve after finding the tensions and everythin..anyway if u still need help in part 2 tell me and i am gonna help ya!

 

[AhShun]

if u knw how to resolve then ur on the safe side haha resolving for those of question is the best

hahaha... but i still leave many mistakes while doing them under exam conditions