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# M1 AS Level 2015 Paper Question 1 Forces In A Lift watch

1. Is there anyone who can help me please? I'm doing an M1 past paper from Summer 2015 and I'm struggling to answer the first question correctly.
The question is;

There is a man, mass Mkg, standing on the floor of a lift that is moving with the acceleration of 0.2ms^-2. The reaction between the floor of the lift and the man is 680N. The mass of the lift is 1800kg. Discover the value of M and the tension in the cable of the lift.

I've tried using F=ma for both questions but I seem to be getting strange answers that don't make sense, for the tension I got 18,816N and for the mass I got -71kg which I know isn't right. I may have just attempted the question wrong but any help will be massively appreciated, thank you!!
2. F=ma
680=mg
m=680/9.8kg
T=(m+m)(a+a)
T=(1800+(680/9.8))*(9.8+0.2)=18694N
3. F in the formula F=ma is the net force. That means that the net force is 0.2M newtons, where M is the man's mass.
Does it say the lift is accelerating upwards or downwards?
4. (Original post by B_9710)
F in the formula F=ma is the net force. That means that the net force is 0.2M newtons, where M is the man's mass.
Does it say the lift is accelerating upwards or downwards?
It is accelerating upwards
F=ma
680=mg
m=680/9.8kg
T=(m+m)(a+a)
T=(1800+(680/9.8))*(9.8+0.2)=18694N
You seem to be thinking that by Newtons 3rd law, the weight of the man is equal to the reaction force. This is NOT true.
6. Moved to maths.
7. (Original post by B_9710)
You seem to be thinking that by Newtons 3rd law, the weight of the man is equal to the reaction force. This is NOT true.
Huh? It isn't?
8. (Original post by Zacken)
Moved to maths.
9. (Original post by abbiechantellex)
It is accelerating upwards
Consider the forces only on the man for the first part of the question - to find the mass of the man.
Huh? It isn't?
This is only true when the acceleration of the lift is 0.
11. (Original post by B_9710)
Consider the forces only on the man for the first part of the question - to find the mass of the man.

Am i right to say the only forces on the man are 680N up and Mg down? But then what would they be equal to?
Sorry, I'm not the best at Mechanics haha
12. (Original post by B_9710)
This is only true when the acceleration of the lift is 0.
13. (Original post by abbiechantellex)
Am i right to say the only forces on the man are 680N up and Mg down? But then what would they be equal to?
Sorry, I'm not the best at Mechanics haha
Resolve upwards, that gets you 680 - Mg = M(a) where a is your acceleration by using Newton's second law. I'm just posting this because B_9710 seems to have gone missing for a sec, I haven't even looked at the question so this might be wrong.
14. (Original post by abbiechantellex)
Am i right to say the only forces on the man are 680N up and Mg down? But then what would they be equal to?
Sorry, I'm not the best at Mechanics haha
Yes, as for what this is equal to, think about it. If the lift is accelerating at 0.2ms^-2 then the man is also accelerating at 0.2ms^-2. So it is equal to 0.2M newtons.
15. (Original post by Zacken)
Resolve upwards, that gets you 680 - Mg = M(a) where a is your acceleration by using Newton's second law. I'm just posting this because B_9710 seems to have gone missing for a sec, I haven't even looked at the question so this might be wrong.
That's the paper.

I gave your suggestion a go, this is what I got.

680 - 9.8m = 0.2m
680 = 10m
m=68

Which makes alot more sense now
16. (Original post by B_9710)
Yes, as for what this is equal to, think about it. If the lift is accelerating at 0.2ms^-2 then the man is also accelerating at 0.2ms^-2. So it is equal to 0.2M newtons.
That makes so much more sense now, thank you!! I got the answer of 68kg which I think is right

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