Lift accelerating upward with passenger on weighing scale

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Klejdi90
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Homework Statement


g= 9.8 Nkg-1

Q. A lift has a mass of 400 kg. A man of mass 70 kg stands on a weighing machine fixed to the floor of the lift. Four seconds after starting from rest the lift has reached its maximum speed and has risen 5 m.

a) What will be the reading on the weighing machine during the point of acceleration?

Homework Equations


A= acceleration
S=distance (5m)
t=Time

A=s/t,F=m*a

The Attempt at a Solution

a = 1.25ms^-1 unsure?

I am really stuck

How would i go about obtaining the change of weight on the scale?
 
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Welcome to PF!

Hi Klejdi90! Welcome to PF! :smile:

(try using the X2 icon just above the Reply box :wink:)
Klejdi90 said:
a = 1.25ms^-1 unsure?

(ms-1 would be a speed)

no, that would be for a speed of 5, this is a distance …

use one of the other standard constant acceleration equations …

what do you get? :smile:​

How would i go about obtaining the change of weight on the scale?

Use F = ma :wink:
 
?I think this may be it:

S=ut+1/2at2

S=5
U=0
t=4

so 5= 0x4+0.5xax42
a= 5/ 0x4+0.5x42
a=0.625m/s2and for the change in mass

F=ma

so F=75x9.8^-1

mass of man on weighing scale = 75 kg
 
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Hi Klejdi90! :smile:
Klejdi90 said:
a=0.625m/s2


Yup! :biggrin:
and for the change in mass

F=ma

so F=75x0.625

which comes to 43.75

There are two forces on the man: mg and N.

N is the same as the weight shown on the weighing machine.

Put them into F = ma, to find N. :smile:
 
tiny-tim said:
Hi Klejdi90! :smile:



Yup! :biggrin:


There are two forces on the man: mg and N.

N is the same as the weight shown on the weighing machine.

Put them into F = ma, to find N. :smile:

sorry to say I'm a little confused :(

F= m*g*?
 
No, "F" in good ol' Newton's "F = ma" is always the total force. :smile:
 
I was wondering if the acceleration was uniform in this question? Why?

Plus would i need to apply the total mass lift and the mans weight so, 400kg+70kg?

There is and upward force due to acceleration and downward force due to g.

So do i do F = ma
F = 70x0.625
F = 43.75N at this point i have the upward force?

The downforce?

Sorry I am very confused I am doing a course which lasts 1 year and teaches A level equivelant physics, plus my teacher is very fast in explaining. I have to give this assignment in tomorrow:(
 
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Hi Klejdi90! :smile:
Klejdi90 said:
I was wondering if the acceleration was uniform in this question? Why?

The question doesn't say so (it should), but I expect you're meant to assume so. :wink:
So, F=mga?

erm :redface: … that doesn't even make sense …

how can you multiply two accelerations? :confused:
Plus would i need to apply the total mass lift and the mans weight so, 400kg+70kg?

read the question …
Klejdi90 said:
Q. A lift has a mass of 400 kg. A man of mass 70 kg stands on a weighing machine fixed to the floor of the lift. Four seconds after starting from rest the lift has reached its maximum speed and has risen 5 m.

a) What will be the reading on the weighing machine during the point of acceleration?

what is the reading on the weighing machine equal to? :smile:
 
Sorry, I meant physically what is the reading on the weighing machine equal to?

(ie, what force does the machine measure? :wink:)
 
That's right … it measures the reaction force between the man and the weighing machine …

by good ol' Newton's third law, the force of the man on the weighing machine is equal (but opposite) to the force of the weighing machine on the man …

so your Ftotal = ma has nothing to do with the lift, does it? :smile:
 

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