physics homework assignment 2

8/10/2019 Physics Homework Assignment 2

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Sumer Vaid

Professor Birons Physics 13100

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Physics Homework Assignment 2

Question 1

i) By Newtons Second Law:

(

ii)

Therefore, by Newtons second law:

Hence, According to Newtons Third Law, let Therefore,

Therefore,

Also,

Hence,

Finally,

At upper edge of the rope:

At the bottom edge of the rope:


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Question 2:

a) Four scenarios:

If the y-acceleration is positive (if the elevator is accelerating upwards)

then: The normal force exerted by the elevator floor on the weighing

scale increases, which then exerts a greater force on the person.Therefore, FN =mg+ma, hence the apparent weight on the weighing scale

will appear greater.

If the y-acceleration is negative (if the elevator is accelerating

downwards) then: The normal force exerted by the elevator on the

weighing scale decreases, which then exerts a lesser force on the person.

Therefore, FN =mg ma, hence the apparent weight on the weighing scale

will appear lesser.

If the elevator is moving downwards and slowing down, the result will be

the same as the case with positive acceleration apparent weight will

appear greater than the true weight. If the elevator is moving upwards and slowing down, the result will be the

same as the case with negative acceleration apparent weight will appear

to be lesser than the true weight.

b) Since the scale and the person will fall together in free fall, there will be

no contact force resulting in the weighing scale showing 0 weight.

Question 3

a) According to the diagram:

Therefore,

Hence, shown

b) The double derivative of displacement with

respect to time is equivalent to the acceleration. Hence:

Hence, shown

c) Diagram 1 (box):

Normal force

Weight = mg


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Diagram 2 (wedge):

d) The only horizontal force acting on the block is the horizontal component

of the normal force, .Therefore,

Also, the vertical component of the normal force, , is actingopposite to the weight of the box The difference between these forcesshould yield the net vertical force acting on the box. Hence:

.For the wedge:

Total vertical force experienced by the wedge = 0 = (no vertical movement of the wedge, hence vertical f=0)

Total horizontal force experienced by the wedge=

, caused

because of the horizontal component of the normal force acting from the box.

From Part B, we also know that

Upon solving this system of equations algebraically, the following expression for

N is obtained:

Therefore, finally:

Normal force exerted by box

Weight = Mg+ mg

Normal force

exerted by floor

(


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e) If M is much bigger than m, the wedge does not displace as a the acceleration

caused by the force calculated in the previous section is negligible.

Question 4

a) Radius of the circular path Circumference of the path = hence,

Also,

Hence,

b)

Weight = mg

R sin


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c)

Hence,

Therefore,

e) I) No the bead cant achieve vertical equilibrium at the centre of the circle

since then the positive force acting on the bead will be completely

horizontal, pointing towards the centre of the loop. Since there will be no

perpendicular vertical force to balance out the weight of the bead, it will

not be able to enter vertical equilibrium.

Mathematically put:At the center of the loop,

This implies that

but this is not possible as g can never be 0.ii) If the ball stays at rest at the bottom:

Therefore,

Hence, if the mathematical condition of

is satisfied, the ball will stay at thebottom.

physics homework assignment 2

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