Vertical and horizontal components,physics As level help :(

[Maryam9]

Homework Statement

This is paper from june 2016 , the link below;
http://qualifications.pearson.com/c...2013/Exam materials/WPH01_01_que_20160524.pdf
Can someone pleeeeeze explain how to solve question 15 part B AND queation 16 part B .
I have an exam tmrw so please please if someone could explain this to me

Homework Equations

How do write the answer in part b of Q16
How to solve the question?[/B]

The Attempt at a Solution

for question 15 I tried using the tension formula T=mg+ma but don't know how to subtitue quantities in it
for question 16 I don't understand what the question is asking about angles , all I could understand that forces cancel out now if its right or no idk but idk how to write the answer :(

 

[BvU]

Hi Maryam,

At the risk of being severely punished by our good mentors (answering is forbidden in PF, we are only allowed to help according to the rules ) - but heck, damsels in distress...

For 15a you may assume the whole lot is just hanging: no acceleration. There's only one fundamental difference between the fly FBD and that for the spider: there is a T₁ pulling down. Once you have the FBDs you also have the equations for equilibrium (= hanging still).
Quantities are the givens (all in the form of symbols).

But you appear to ask about 15b: They give T₁ and you will need to solve TWO equations of the type T=mg+ma : one for the fly and one for the spider. Look at the FBDs and write them down (in terms of the symbols).

For 16: You solved a) ? For b) you could look at T as a function of $\theta$ Or you could look at the horizontal component T$\sin\theta$ because that is what is pulling the pole tips towards each other.

 

[Maryam9]

Hi Maryam,

At the risk of being severely punished by our good mentors (answering is forbidden in PF, we are only allowed to help according to the rules ) - but heck, damsels in distress...

For 15a you may assume the whole lot is just hanging: no acceleration. There's only one fundamental difference between the fly FBD and that for the spider: there is a T₁ pulling down. Once you have the FBDs you also have the equations for equilibrium (= hanging still).
Quantities are the givens (all in the form of symbols).

But you appear to ask about 15b: They give T₁ and you will need to solve TWO equations of the type T=mg+ma : one for the fly and one for the spider. Look at the FBDs and write them down (in terms of the symbols).

For 16: You solved a) ? For b) you could look at T as a function of $\theta$ Or you could look at the horizontal component T$\sin\theta$ because that is what is pulling the pole tips towards each other.

Aww thanks !
What does FBD mean ?? And in question 15 they asked for the acceleration not tension , How do I get that??
And question 16 I was asking about part b , Idk how to write the answer because I don't understand it

 

[BvU]

Free body diagram

T₂ - m_flyg = ma for the fly
T₁ - T₂ - m_spiderg = ma for the spider.

a is the same for both (The distance between them does not change). Eliminate T₂ and there you are !

16: if $\theta$ increases, the horizontal components of T decrease, so the poles are not pulled together so hard any more. The vertical components remain the same.

 

[Maryam9]

Free body diagram

T₂ - m_flyg = ma for the fly
T₁ - T₂ - m_spiderg = ma for the spider.

a is the same for both (The distance between them does not change). Eliminate T₂ and there you are !

16: if $\theta$ increases, the horizontal components of T decrease, so the poles are not pulled together so hard any more. The vertical components remain the same.

ooh! ok Thank u so much :)

 

[BvU]

You're welcome. Good luck with the test.