Mechanics PAPER4 2009/ 41

[1992]

This questions is strange, i still canot do it after many tries :S

Its Q4. i)..... How can i posible find the tension in one string??¿? :S ive decomposed the triangle and everything but i canot get the answer. Working our needed pleaseee

I know how to do part ii) and iii) if I get the answer for part i)

 

[Anomaly]

This questions is strange, i still canot do it after many tries :S

Its Q4. i)..... How can i posible find the tension in one string??¿? :S ive decomposed the triangle and everything but i canot get the answer. Working our needed pleaseee

I know how to do part ii) and iii) if I get the answer for part i)

Resolve both tensions [call them T1(tension in the shorter string) and T2(tension in the longer string)] into vertical and horizontal components. As the system is in equilibrium, the horizontal components of T1 and T2 are equal this gives you an equation T1sin36.9 = T2sin53.6 which simplifies o 0.6T1=0.8T2, now same for the vertical components, those give the equation T1cos36.9+T2cos53.1=5 which simplifies to 0.8T1 +0.6T2=5. You have two equations in T1 and T2, solve to get T2.

 

[Anomaly]

Another method is that, you can resolve the 5N force into 2 components which simply leads to 5sin36.9=T2, however in order to find the relevant angles you'll have to work out some angles.

 

[1992]

ok thanks
but how do u know the tension is diferent. for eg: in O/N 2002: Q3, the tension are the same althouhg it have diferent lenght.
that is waht i got confused.