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If lift at rest or moving with constant velocity,the reaction R will be equal to the weight ofthe student, W
R=W
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REACTION FORCE WILL BE MORE THAN THEWEIGHT OF THE STUDENT, W.
F=R-W
ma=R-mgR=ma+mg
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The reaction force R will be less than theweight of the student, W.
F=W-R
ma=mg-R R=mg-ma
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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)
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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 )
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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
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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)
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(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
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(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
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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
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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?
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(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
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(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.
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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
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(a)Resultant force,F=W-f
=5(10)-10
=40N
(b) F=ma
a= F = 40
m 5+3=5ms-2
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(c) tension of the string
F=T-f
ma=T-fT=ma +f
=3(5) +10
=25N
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