Vertical and horizontal components,physics As level help :(

In summary: Remember, the angles are in degrees, not radians.The conversation is about solving questions 15 and 16, parts B for an exam. The first question involves using the tension formula to solve for quantities, while the second question involves finding the horizontal component of T. The conversation also includes a discussion on free body diagrams and the effects of changing angles on tension.
  • #1
Maryam9
3
0

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 :cry:

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 :(
 
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  • #2
Hi Maryam, :welcome:

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.
 
  • #3
BvU said:
Hi Maryam, :welcome:

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 !:-p
What does FBD mean ?? And in question 15 they asked for the acceleration not tension , How do I get that?? :frown:
And question 16 I was asking about part b , Idk how to write the answer because I don't understand it
 
  • #4
Free body diagram

T2 - mflyg = ma for the fly
T1 - T2 - mspiderg = 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.
 
  • #5
BvU said:
Free body diagram

T2 - mflyg = ma for the fly
T1 - T2 - mspiderg = 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 :) :dademyday:
 
  • #6
You're welcome. Good luck with the test.
 
  • Like
Likes Maryam9

Related to Vertical and horizontal components,physics As level help :(

1. What are vertical and horizontal components in physics?

Vertical and horizontal components are two types of vector quantities that describe the direction and magnitude of motion. The vertical component is the motion in the up and down direction, while the horizontal component is the motion in the left and right direction.

2. How are vertical and horizontal components related?

The vertical and horizontal components are related through the use of trigonometric functions. The vertical component can be calculated by multiplying the magnitude of the vector by the sine of the angle it makes with the horizontal axis. Similarly, the horizontal component can be calculated by multiplying the magnitude of the vector by the cosine of the angle.

3. How do vertical and horizontal components affect projectile motion?

In projectile motion, the vertical and horizontal components of the initial velocity determine the path of the projectile. The vertical component causes the object to rise and fall, while the horizontal component causes it to move forward at a constant speed. The two components work together to create a parabolic path.

4. What is the difference between vertical and horizontal components?

The main difference between vertical and horizontal components is the direction in which they act. The vertical component is perpendicular to the horizontal component. Additionally, the vertical component is affected by gravity, while the horizontal component remains constant in the absence of external forces.

5. How do I calculate the vertical and horizontal components of a vector?

To calculate the vertical and horizontal components of a vector, you can use trigonometric functions such as sine and cosine. First, determine the angle the vector makes with the horizontal axis. Then, multiply the magnitude of the vector by the sine of the angle to find the vertical component and by the cosine of the angle to find the horizontal component.

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