apparent weight in a lift

7/30/2019 Apparent Weight in a Lift

1/18


2/18

If lift at rest or moving with constant velocity,the reaction R will be equal to the weight ofthe student, W

R=W


3/18

REACTION FORCE WILL BE MORE THAN THEWEIGHT OF THE STUDENT, W.

F=R-W

ma=R-mgR=ma+mg


4/18

The reaction force R will be less than theweight of the student, W.

F=W-R

ma=mg-R R=mg-ma


5/18

A cat of mass 7.0 kg rests on a weighingmachine in a lift. What is the weight of the catwhen the lift

a) moves upwards with an acceleration of2.0ms-2?

b) Moves downwards with an acceleration of3.0ms-2?

c) Moves downwards with a constant velocity?(use g=10ms-2)


6/18

A mass of 2kg is suspended from the roof ofa lift by a spring balance as shown in Figure2.78. What will the reading on the springbalance be if the lift is accelerating upwards

with a constant acceleration of 1ms-2 ?(g=10ms-2 )


7/18

As the lift is accelerating upwards, theresultant force that acts on the mass will begiven by:

F= R-W

ma= R-mgR=ma +mg=m (a+g)=2(1+10)

=22 NReading on the spring balance= 22N


8/18

A 15 kg box is supported ona rope that passes through apulley. The other end of therope is held by a 45kgstudents standing on aweighing machine as shownin Figure 2.79(a). Assuming

that the pulley is smooth andthe rope is of negligiblemass, what will the readingon the weighing machine bewhen

(a) the box is at rest? (b) the box is accelerates

upwards with an accelerationof 2 ms-2 ?(g=10ms-2)


9/18

(a) As the rope is at rest, the

tension T1 in the rope isequal to the weight of the boxm1g.

(m1=mass of the box)

T1=m1g

=15 x 10=150N

Since the students is also atrest, the sum of normalreaction R1 and the tension inthe rope T1 is equal to theweight of the student m2g.

m2= mass of the student

R1+T1= m2g

R1= m2g-T1= 45 (10) -150 = 300N


10/18

(b) The forces that act on thesystem when the boxaccelerates upwards isshown in Figure 2.79(c). Asthe box acceleratesupwards with anacceleration of 2ms-2, theresultant force on the boxis

F=T2-m1gm1a=T2-m1gT2 = m1a+m1g

=m1 (a+g) =15(2+10)=180N


11/18

Since the student is atrest, the sum of thenormal reaction R2 andthe tension is the rope T2is equal to the weight of

the student m2g.R2+T2 = m2g

R2 = m2g T2=45(10)-180

=270N


12/18

An object of mass 30kg is supported by a force F inthe pulley system shown in Figure 2.68. Assumingthat the pulley us smooth and is negligible mass,find the force F when

(a) The object moves upwards with a constant

velocity(b) The object moves upwards with an acceleration of

2 ms-2?


13/18

(a) When the object moves with a constantvelocity, the forces act on it are inequilibrium, thus the applied force F willequal to the weight of the object W.

F=W

= mg

= 30 x 10

= 300N


14/18

(b) Resultant forceF=F-W

ma=F-W (F=ma)

30(2)=F-300F=360N

The force F=360N when the object acceleration

at 2ms-2.


15/18

A 5 kg mass is used to accelerate a 3kg trolleyalong a table as shown in Figure.The frictionbetween the table and the trolley is10N.Assuming that the pulley is smooth and

the string is negligible,find(a) The resultant force that acts on the system

(b) The acceleration on the system

(c) The tension in the string


16/18


17/18

(a)Resultant force,F=W-f

=5(10)-10

=40N

(b) F=ma

a= F = 40

m 5+3=5ms-2


18/18

(c) tension of the string

F=T-f

ma=T-fT=ma +f

=3(5) +10

=25N

apparent weight in a lift

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