M1 - Can any1 help pleeeaaaaassssseeeeee!?!?!?!?! Watch

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sam1111_9
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Hey all, i don't like asking 4 help like this, but i have this really bad teacher who never tells us how to do anything, and just gives us questions to do, if we can't do them she goes mental and gives us detention. So if any one can help with these questions it would be so much appreciated!

5. A fork-lift truck is raising a container of car batteries with an acceleration of 1.2ms-2. The normal contact force on the container from the horizontal forks is 1485N. Calculate the mass of the load.

7. The pilot of a hot air balloon has mass 85kg. As the balloon leaves the ground, the normal contact force on the pilot from the floor of the balloon immediately increases to 901N and remains at this value for the first stage of the ascent. Calculate the acceleration of the balloon in this stage of its motion.

9. A lift starting from rest moves downwards with constant acceleration. It covers a distance s in time t , where s= 1/6 gt squared. A box of mass m is on the floor in the lift. Find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.

11. A man of mass M kg and his son of mass m kg are standing in a lift. When the lift is accelerating upwards with magnitude 1.2ms-2 the magnitude of the normal contact force exerted on the man by the lift floor is 880N. When the lift is moving with constant speed the combined magnitude of the normal contact forces exerted on the man and the boy by the lift floor is 980N. Find the values of M and m.

If you are able to do any of these, then it means i will dare go back to school. Thankyou all soooooooooo much!!!
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kikzen
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right, ive just done m1 too but i havent really revised it yet (after the holidays yeh?) so im not sure if these are right

5. Resultant force = mass x accel
1485 - mg = m(1.2)
1485 = m(1.2) + mg
1485 / (1.2 + g) = m
m = 135Kg

7. again using the same rule
901 - 85g = 85a
a = 901/85 - g
a = 0.8ms^-2

9. not sure about tbh but i think this is it
s = 1/6gt^2
you have u = 0
t = t
a = ?
s = ut + 1/2at^2
1/6gt^2 = 0 + 1/2at^2
t^2 cancels
divide by 1/2
1/3g = a

sub this back into the lift
R - m = 1/3g(m)
R = 1/3gm + m
R = m(1/3g + 1)

11. use the same way as 7., i think you need to deal with when the lift is accelerating and when its moving constantly (acceleration = 0). you may be able to take the components individually (ie man and child separaretly). probably.
(accelerating downwards : 880 - Mg = 1.2M
880 / (1.2 + g) = M = 80Kg

constant accel : 980 - 80g - mg = 0
980/g - 80 = m = 20Kg)
that right?

good luck!
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crana
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Hey
Just having a quick look through a couple of these and adding comments to help you rather than doing them for you..hope they aren't wrong... I'm assuming the accelerations given are resultant.

(Original post by sam1111_9)
5. A fork-lift truck is raising a container of car batteries with an acceleration of 1.2ms-2. The normal contact force on the container from the horizontal forks is 1485N. Calculate the mass of the load.
Use F=ma
1485 = 1.2 m

(Original post by sam1111_9)
7. The pilot of a hot air balloon has mass 85kg. As the balloon leaves the ground, the normal contact force on the pilot from the floor of the balloon immediately increases to 901N and remains at this value for the first stage of the ascent. Calculate the acceleration of the balloon in this stage of its motion.
.
F = ma
901 = 85 a

(Original post by sam1111_9)
9. A lift starting from rest moves downwards with constant acceleration. It covers a distance s in time t , where s= 1/6 gt squared. A box of mass m is on the floor in the lift. Find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.
.
Use s = ut + 1/2 at^2 to find acceleration in terms of g. Then use F=ma.

(Original post by sam1111_9)
11. A man of mass M kg and his son of mass m kg are standing in a lift. When the lift is accelerating upwards with magnitude 1.2ms-2 the magnitude of the normal contact force exerted on the man by the lift floor is 880N. When the lift is moving with constant speed the combined magnitude of the normal contact forces exerted on the man and the boy by the lift floor is 980N. Find the values of M and m.

1) Use F=ma for the man (F=880, a = 1.2) to find M

2) At constant speed, a(from the lift) =0. So, only acceleration due to gravity (g) is relevant (9.8 or 9.81 depending how picky your teacher is). Use F=mg to find the total mass of man and boy (M+m) then subtract the value from step 1 to find m .

Hope this helps, you can PM me with any probs
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sarahwhatevver
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(Original post by sam1111_9)
Hey all, i don't like asking 4 help like this, but i have this really bad teacher who never tells us how to do anything, and just gives us questions to do, if we can't do them she goes mental and gives us detention. So if any one can help with these questions it would be so much appreciated!

9. A lift starting from rest moves downwards with constant acceleration. It covers a distance s in time t , where s= 1/6 gt squared. A box of mass m is on the floor in the lift. Find, in terms of m and g, an expression for the normal contact force on the box from the lift floor.



If you are able to do any of these, then it means i will dare go back to school. Thankyou all soooooooooo much!!!
I am not sure about this but it is a possibility....

#9
s=1/6 gt(squared)
s=ut + 1/2 at(squared)
set both equal and u=0
thus, a=1/3g
F-mg=ma
plug in 'a' in the equation and you get F=mg(1/3 + 1)
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sarahwhatevver
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(Original post by sam1111_9)
Hey all, i don't like asking 4 help like this, but i have this really bad teacher who never tells us how to do anything, and just gives us questions to do, if we can't do them she goes mental and gives us detention. So if any one can help with these questions it would be so much appreciated!


If you are able to do any of these, then it means i will dare go back to school. Thankyou all soooooooooo much!!!
This may seem silly but....what is M1?
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kikzen
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(Original post by sarahwhatevver)
This may seem silly but....what is M1?
mechanics 1 (a level module)
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