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Mathematics: Post your doubts here!

hellodjfos;s'ff

hellodjfos;s'ff

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Mistar-Kigi said:
Could you tell me that total energy transformed and value of Resistance of wire you got? And also the time you calculated in the last part of waves question.
Thanks!
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12 s for last part of waves question. Total energy transformed was 6000 J which can be found through the formula, E=IVt(you would have learnt this in olevels).
 
amina1300

amina1300

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haithsoul said:
rearrange the equation to cos4x-4cos2x=3 now use the identity from part i to end up with
8sin^4x-3=3 now you can solve this part like usual. 8sin^4x=6 then sin^4x=6/8 then get x.
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Oh, the 3 in the second given equation!! Im so dumb :cry::cry:
 
N

Nushad Nahue

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Capture.PNG Aright guys, M2, s05 Q3. part 2.

It says here :
The solid rests in equilibrium with the face containing the edge AF of the cross-section in contact with a horizontal table. The weight of the solid is W N. A horizontal force of magnitude P N is applied to the solid at the point B, in the direction of BC (see Fig. 2). ( I have uploaded the figure)
The table is sufficiently rough to prevent sliding.

(ii) Find P in terms of W, given that the equilibrium of the solid is about to be broken.

In the figure it is clear that I have to take the moment about F.

However in the mark scheme it is given: For obtaining an equation in P and W by taking moments about F and using the idea that the normal component of the contact force has no moment about F (almost certainly implied in most cases).
I get that from F, you get an anticlockwise moment of the weight and a clockwise moment of the force P N. However, how can the normal component of the contact force have zero moment about F??? It acts opposite to the direction of W so shouldn't it have a clock-wise moment along with P?
 
D

DudeWithoutName

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Nushad Nahue said:
View attachment 63616 Aright guys, M2, s05 Q3. part 2.

It says here :
The solid rests in equilibrium with the face containing the edge AF of the cross-section in contact with a horizontal table. The weight of the solid is W N. A horizontal force of magnitude P N is applied to the solid at the point B, in the direction of BC (see Fig. 2). ( I have uploaded the figure)
The table is sufficiently rough to prevent sliding.

(ii) Find P in terms of W, given that the equilibrium of the solid is about to be broken.

In the figure it is clear that I have to take the moment about F.

However in the mark scheme it is given: For obtaining an equation in P and W by taking moments about F and using the idea that the normal component of the contact force has no moment about F (almost certainly implied in most cases).
I get that from F, you get an anticlockwise moment of the weight and a clockwise moment of the force P N. However, how can the normal component of the contact force have zero moment about F??? It acts opposite to the direction of W so shouldn't it have a clock-wise moment along with P?
Click to expand...
Sure, if you want to calculate the moment of the normal component of the contact force about F then by using Moment=Force*Perpendicular distance of force from pivot=N*0=0Nm because if it acts at F, its distance from F is 0 (Hope that made sense)
 
D

DudeWithoutName

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Nushad Nahue said:
Also, M2, w04, Q7 part 2.

the mark scheme says: Centre of mass of triangle is 3 2y from interface B1 For using ‘moment about the interface = 0’ or equivalent or ‘moment about AD = 0.4A’ AND with areas in terms of y M1 [0.4y × 0.2 = ½ 2 y × y × 3 2y or 0.4(0.4y + ½ 2y × y) = 0.4y × 0.2 + ½2y × y (0.4 + 3 2y )]

I get the For using ‘moment about the interface = 0’ or equivalent part. BUT,

THIS PART
moment about AD = 0.4A’ AND with areas in terms of y \ 0.4(0.4y + ½ 2y × y) = 0.4y × 0.2 + ½2y × y (0.4 + 3 2y )]
So the forces acting on the container is the weight of mass of the rectangular part and the weight of the triangular part and the contact force?
I am not sure but I assume that the LHS of the equation comes from the contact force?? I am not sure. If so then, I think I am confused about the direction of the moment. I don't know where and how the LHS equation came about and if it is the contact force, I am confused about the direction.View attachment 63617 Can someone please explain it to me?
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Hmm, when it is on the point of slipping, the contact force will be on B, so if we take B as the pivot, we make the distance between the contact force and the pivot 0 hence cancelling its moment, right? So now it'll be much more convenient. We calculate the moment of the triangle and the rectangle and set them equal to each other then solve for y, hence finding the critical value of y for which the container is about to slip. Moment of rectangle: 0.4y*0.2. Moment of triangle: 0.5*2y*y*(1/3)(2y)=(2/3)y^3
(2/3)y^3=0.08y.....y=0.346
 
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