You are Here: Home >< Maths

# MEI Mechanics 1- questions watch

1. http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
Also for question 7vi, i don't get whats going on

Thank you
2. (Original post by runny4)
http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
Uh, you can? I think - it's a fine argument. There's a component of force acting at right-angles to the motion, hence the resultant force will not be in the direction of the motion.

Also for question 7vi, i don't get whats going on

Thank you
The part of the string BC is 2.3 metres which is greater than 2 metres, i.e: it is "too long" and the piece BC will be effectively useless. Imagine holding a block using a chain/string that is 2 metres long when the block is situated one metre away from you, the chain/string will be pretty much useless, it'll be just limp (slack) and doing nothing. That's the case here: the piece AB will have to handle the entirety of the weight (50N) and you'll find that the angle CAB is 90 degrees, so it's basically just resolve upwards/downwards to get AC = 50 N and BC = 0 / useless.
3. (Original post by Zacken)
Uh, you can? I think - it's a fine argument. There's a component of force acting at right-angles to the motion, hence the resultant force will not be in the direction of the motion.

The part of the string BC is 2.3 metres which is greater than 2 metres, i.e: it is "too long" and the piece BC will be effectively useless. Imagine holding a block using a chain/string that is 2 metres long when the block is situated one metre away from you, the chain/string will be pretty much useless, it'll be just limp (slack) and doing nothing. That's the case here: the piece AB will have to handle the entirety of the weight (50N) and you'll find that the angle CAB is 90 degrees, so it's basically just resolve upwards/downwards to get AC = 50 N and BC = 0 / useless.
Just for the 2nd question if you use the cosine rule to find angle CAB, you get
(4+0.25-2.3^2)/(2*2*0.5)= cos(CAB) but this does not give you CAB=90
4. (Original post by runny4)
Just for the 2nd question if you use the cosine rule to find angle CAB, you get
(4+0.25-2.3^2)/(2*2*0.5)= cos(CAB) but this does not give you CAB=90
Right, but if you're using the cosine rule like that, then you're assuming BC is taut when it isn't.
5. (Original post by Zacken)
Right, but if you're using the cosine rule like that, then you're assuming BC is taut when it isn't.
Then how would you deduce that Angle CAB is 90 degrees?
6. (Original post by runny4)
http://www.mei.org.uk/files/papers/M1_Paper_Jan13.pdf
For the above above paper question 3iii, why couldn't you just resolve perpendicular to motion and say that there is a resultant force because Sam's and Tom's components aren't the same at right angles to motion. (the mark scheme is at the bottom of the pdf)
IMO, saying there is a resultant force perpendicular to the acceleration isn't sufficient. There must be something counteracting that resultant force for the car to accelerate as it does. I.e. a resistive force perpendicular to the motion.
7. (Original post by ghostwalker)
IMO, saying there is a resultant force perpendicular to the acceleration isn't sufficient. There must be something counteracting that resultant force for the car to accelerate as it does. I.e. a resistive force perpendicular to the motion.
but its asking why the resultant of the forces exerted by Sam and Tom is not in the direction of the car’s acceleration not why the car travels in a straight line
8. (Original post by runny4)
but its asking why the resultant of the forces exerted by Sam and Tom is not in the direction of the car’s acceleration not why the car travels in a straight line
it's not about why the car travels in a straight line.

Your answer is only showing that the sum of the perpendicular components of the forces (of the two people) is non-zero. This does not explain why the car moves in the direction it does.

The car will accelerate in the direction of the sum of the forces acting on it - always. The question is asking why is this direction different to the resultant of the forces exerted by the two people.
9. (Original post by ghostwalker)
it's not about why the car travels in a straight line.

Your answer is only showing that the sum of the perpendicular components of the forces (of the two people) is non-zero. This does not explain why the car moves in the direction it does.

The car will accelerate in the direction of the sum of the forces acting on it - always. The question is asking why is this direction different to the resultant of the forces exerted by the two people.
ok thanks

### Related university courses

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: April 1, 2016
Today on TSR

### He lied about his age

Thought he was 19... really he's 14

### University open days

Wed, 25 Jul '18
2. University of Buckingham
Wed, 25 Jul '18
3. Bournemouth University
Wed, 1 Aug '18
Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams