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Can someone please help me in JUN 2012 P11 Qs.( 5 , 8 , 15 , 37)
JUN 13 P12 Qs.( 6 , 20 , 32)
NOV 13 P13 Qs. ( 31 , 32 , 36 , 37 , 38 )
I know its a lot of questions but ive been so confused
Your help would be much appreciated.
Q15)
We have to consider the two elements or conditions of equilibrium:
i) For translational equilibrium, no net force must act on the system;
ii)For rotational equilibrium, not net torque must act on the system about any point.
Let's take the crane itself to be the system. Then forces are exerted on the crane by the Earth (Gravity), the weight W (this is stationary so it has to be balanced), the weight L (same justification as previous one) and the ground (Normal force, maybe frictional force).
Now the first thing to note is that moving the weight L to the right, the crane feels a tendency to rotate in the clockwise direction, since the net moment about the center (Let's take that point to be at the top of the vertical rod, and at it's middle) will try to rotate it in that direction.
So if the crane tries to rotate clockwise, for it to be stationary there has to be some other torque that tries to rotate it anti-clockwise, right?
This is the second part of the conditions, and yes, there has to be a torque that tries to rotate the system anticlockwise.
This eliminates A - the force by the ground on A will only rotate it further clockwise, not resist the clockwise motion (imagine the crane is a door hinged at the top, and apply that force - the door will rotate clockwise, right? This is the opposite of what we want happening).
B is possible - the force by the ground attempts to rotate the "door" anti-clockwise and could thus aid in stabilizing the system, but we'll get back to that later.
C is wrong - suppose the normal force R moves to the left, then it will still try to rotate the system in a clockwise manner (since it is not perfectly aligned with that point we are taking moments about, it will try to rotate the system clockwise, which is not what we want).
D is possible - the reaction force R attempts to turn the crane anticlockwise and thus could stabilize the system, but there's a little more work to do.
We have to apply the first condition of equilibrium - that no net force acts on the system.
In B, there is a vertical force of gravity, a vertical force from W, a vertical force from L, a vertical force from the ground and a horizontal force from the ground.
Hang on. This isn't balanced! There is not horizontal force counteracting the horizontal force from the ground, so how can the system remain in equilibrium?
Let's confirm with D. In D, there is no horizontal force that upsets the equilibrium, but there is a variable vertical force that ensures equilibrium both, for translation and rotation. Therefore, D is our answer.
Q37)
Let's put aside the light factor and deal exclusively with the resistance.
First things first - the output voltage will be equal to the potential difference across the resistor R, since the output terminals are connected across R, and a voltmeter connected in the same manner would give the potential difference across R which is what we get in the output circuit.
Since the same current passes through both the resistors, the net resistance will be equal to
(Resistance of R) + (Resistance of LDR)
And since we can write V = IR, we can say
V = 10 - 0 = 10 Volts = I (Resistance of R + Resistance of LDR)
and therefore, I = 10/([Resistance of R] + [Resistance of LDR])
This is the current flowing through both the resistors. Now, we can find out the potential difference across resistance R by again using V = IR to get
V(across R) = Output Voltage = 10 * [Resistance of R]/([Resistance of R] + [Resistance of LDR])
Right now, I suggest you write this down if possible, in whatever manner is easiest for you to understand, since the calculation we can do now is going to be difficult to display here in a post.
Once you're done, note that we have to maximize this value of output voltage. We can do this with calculus, or we can do this with a neat math trick, which i''ll describe below:
We can divide both the numerator and denominator of the Output Voltage equation by [Resistance of R] - this is allowed since while doing this we are essentially multiplying the equation by (1/[Resistance of R])/(1/[Resistance of R]) - this is equal to 1, and multiplying an equation by 1 doesn't change it.
In this post, i'll do it step by step. First let's divide the top half by [Resistance of R] and we get
10 * ([Resistance of R]/[Resistance of R])/([Resistance of R] + [Resistance of LDR]) = 10/([Resistance of R] + [Resistance of LDR])
Now let's divide the bottom half by [Resistance of R] to get
10/([Resistance of R]/[Resistance of R] + [Resistance of LDR]/[Resistance of R]) = 10/(1 + [Resistance of LDR]/[Resistance of R])
So the output voltage is also equal to 10/(1 + [Resistance of LDR]/[Resistance of R]). To maximize this, we have to minimize the denominator, since the numerator is constant and cannot be changed.
Furthermore, since the denominator is equal to (1 + [Resistance of LDR]/[Resistance of R]) we basically have to minimize the second half, since the first half is a constant (1) and cannot be changed.
So, to minimize [Resistance of LDR]/[Resistance of R], we have to maximize [Resistance of R] and minimize [Resistance of LDR], since that will give us the smallest fraction possible.
To minimize the [Resistance of LDR], we note that we have to maximize the intensity of the light falling on the LDR, and so the only option that shows both these options is A.
Hope this helped!
Good Luck for all your exams!
