Thursday, January 8, 2015

Physics 9702 Doubts | Help Page 44

  • Physics 9702 Doubts | Help Page 44




Question 267: [Forces > Moments]
Uniform ladder rests against vertical wall where there is negligible friction. The bottom of ladder rests on rough ground where there is friction. The top of ladder is at a height h above the ground and the foot of the ladder is at a distance 2a from the wall.
The diagram shows the forces which act on the ladder.

Which equation is formed by taking moments?
A W a + F h = 2W a
B F a + W a = F h
C W a + 2W a = F h
D W a – 2W a = 2F h

Reference: Past Exam Paper – June 2008 Paper 1 Q14 & November 2014 Paper 11 & 12 Q12



Solution 267:
Answer: A.
The key to this question is first to find a suitable point about which to take moments.

Since the ladder is not moving (nor rotating), it is in a state of equilibrium.
For equilibrium, the sum of clockwise moments is equal to the sum of anticlockwise moments. This is true for the resultant moment about any point in the system.

The moment of a force is the product of the force and the perpendicular distance of its line of action to the pivot.
So, it is easier to look for a point (to consider as pivot about which we will be considering the moments) where the lines of action of the forces present are perpendicular to it. Otherwise we would need to perform too many calculations (of angles, distances …) which is not appropriate for an MCQ.

Moments about the top of the ladder would give the required equation directly. There is no need to involve any angles when the forces are all given. However here, we will consider the bottom-left corner.

Consider the bottom-left corner where the vertical wall and the ladder meet. The distances are already available and most forces act perpendicularly to a line joining the forces and the point (except the force F at the ground).

Consider the clockwise moments:
The force F (at the top) at a height h from the ground causes a clockwise moment (= Fh).
The weight W of the ladder acts in such a way that the perpendicular distance joining the force and the corner is a. This causes a clockwise moment (= Wa).
Sum of clockwise moments = Wa + Fh

Consider the anticlockwise moments:
The ground causes a reaction W upwards on the ladder, which is a distance 2a from the corner. This causes an anticlockwise moment (= W(2a) = 2Wa). This is the only force that causes an anticlockwise moment.

Note that the line of action of the force F at the ground is not perpendicular to the force. Therefore, this force does not produce any moment.


For equilibrium, Wa + Fh = 2Wa









Question 268: [Forces > Equilibrium of forces]
(a) Fig illustrates a force of 180N acting at a perpendicular distance of 40cm from a point O.

Calculate the moment of the force about O.                          [2]

(b) Fig illustrates a circular drum of radius 8.0cm around which a rope is wound. The drum is turned by means of a handle of length 40cm to which a force of 180N is applied at right angles to the handle.

Calculate the tension T in the rope.                                        [2]

(c) A person is pulling a boat on to a trailer using the drum system from (b), known as a winch, as shown in Fig. The person is exerting a force of 180N on the handle.

(i) Explain why the force the rope exerts on the boat may be less in practice than the tension you have calculated in (b)
(ii) Calculate the force which the trailer exerts on the axle of the winch when the handle is in the position shown in Fig. State the direction in which this force acts.                    [3]

Reference: Past Exam Paper – N97 / II / 2



Solution 268:
(a) Moment = 180 x 0.40 = 72Nm

(b)
For the drum to rotate at constant angular velocity, the tension is the rope produces a moment equal in magnitude to the moment produced by the handle.
{Angular acceleration is zero. Note that similar to the vector F = ma, the vector Torque = (moment of inertia) x (angular acceleration). This is not is the A-level syllabus however.}

The 180N force acts at a distance of 0.40m from the centre of the drum and the tension in the rope acts at a distance of 0.08m from the centre.
T (0.08) = 180 (0.40)
Tension T in the rope = (180 x 0.40) / 0.08 = 900N
{Observe that the length of the winch’s handle and the radius of the drum can provide a significant mechanical advantage. A drum of smaller radius allows a person to more easily turn the handle, but to draw the same length of rope the handle needs to be turned more times.}

(c)
(i)
The calculations in (b) are for an ideal winch. For a real winch, the axle of the drum is subjected to 3 torques. For the axle to rotate at constant angular velocity, the sum of these torques should be zero. The 3 torques are related as follows:
Input torque = Torque due to friction + Output torque
The 180N force of the person produces an input torque, but since a torque is also present due to friction, the output torque is reduced.  
In practice, friction would also cause a torque, reducing the output torque.

(ii)
{Assume that the rope is drawing the boat at a constant velocity. So, the net force is force.}
From Newton’s third law, the following can be concluded:
Since the tension T (= 900N) in the rope acts towards the right {the rope is wrapped around the axle, so this force of 900N is exerted by the axle on the trailer}, there will be a reaction force of equal magnitude by the trailer on the axle of the winch in the opposite direction (to the left). {Physically, this can be understood as follows. A force of 900N acts towards the right on the axle, but the axle itself is not moving forward [there is no linear motion]. So, there must be a force in the opposite direction such that the resultant force on the axle is zero.}
Similarly, a force of 180N is exerted towards the right on the handle which is connected to the axle. So, there should a force equal in magnitude and opposite in direction (towards the left). {The axle is in contact with the trailer through bearings, … and so, exerts a force of 180N on the trailer. Therefore, the trailer also exerts a force of 180N in opposite direction on the axle.}

Total force on axle by trailer = 900 + 180 = 1080 N to the left









Question 269: [Kinematics > Linear motion]
A man stands on the edge of a cliff and throws a stone vertically upwards at 15m.s-1. After what time will the stone hit the ground 20m below?

Reference: ???



Solution 269:
In this question, we are dealing with a 1-dimensional problem with constant acceleration (acceleration of free fall, g = 9.81ms-2). So, the equations for uniformly-accelerated motion may be used.

The key here is to understand that the equations for uniformly-accelerated motion are VECTOR equations. So, the values of the vector quantity involved may be either positive or negative, depending on how the positive direction has been defined.

For the problem, we need to find the time, t for the stone to hit the ground 20m below.
Taking the edge of the cliff as the origin and defining the positive direction being vertically upwards, the DISPLACEMENT (if you are having problems, review the definition of displacement – displacement may be positive or negative, while distance is always positive) is -20m.

So, the initial speed, u = +15ms-1 and the acceleration of free fall, g = -9.81ms-2. We now search for an equation involving these quantities.

Consider the equation for uniformly accelerated motion: s = ut + ½ at2
-20 = 15(t) + 0.5(-9.81)t2

Simplifying and re-arranging gives 4.905t2 – 15t – 20 = 0

This can be solved by the expression for quadratic equation
x = [-b ± √(b2 – 4ac)] / 2a
where a = 4.905, b = -15, c = -20 and x = t.

Solving gives either t = 4.062 (for the + case) or t = -1.004 (for the – case)
Since time cannot be negative, the second solution is neglected (since it is physical impossible)
The solution is therefore, time t = 4.062s









Question 270: [Electric field > Potential]
Refer to November 2007 Paper 4 Q4 (c) (ii) at http://physics-ref.blogspot.com/2014/09/9702-november-2007-paper-4-worked.html

Reference: Past Exam Paper – November 2007 Paper 4 Q4 (c) (ii)



Solution 270:
{The charge induced on the inside of the metal box should have the opposite sign to that of the sphere.}
A negative charge is induced on (the inside of) the box.
{The electric potential at a point is defined as the work done per unit positive charge in moving a small test charge from infinity to the point. At infinity, the electric potential is defined to be zero.

So, energy is required to move the unit positive charge. But why is energy required? 

It is known that like charges repel. Since the sphere is positively charged, it tends to repel the unit positive charge so energy is required to bring the unit positive charge towards the sphere. If the sphere was isolated, the energy (per unit charge) would be (= electric potential) 285JC-1 as calculated before.

EITHER
The formula {V = Q / 4πϵor} applies to isolated (point) charge
OR
{But in this case, as stated, the inside of the metal box is negatively charged. Unlike charges attract. So, the unit positive charge is attracted towards the inside of the metal box. 

That is, 2 forces act on the unit positive charge being considered:
(i) a repulsive force due to the positively charged sphere
(ii) and an attractive force due to the negatively charged inside of the metal box. 

So, the energy required to bring the unit positive charge to the surface of the sphere is less when both forces are present than it would be when only the first repulsive force would have been present. }

Less work is done in moving the test charge from infinity.   

{So, since the unit positive charge is also being attracted by the inside of the metal box, less energy is needed to move it from infinity to the surface of the sphere. Again, the electric potential at a point is the work done (energy required) …..}
So the potential is lower.






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