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Can someone show me how to do Question 29 and 30 from Review Exercise 1 of the Heinmann M2 book. I'm following the method my teacher gave me but its coming out wrong. It would be great if someone could show me how to do it, with a good method and the proper way to lay out and approach these questions.

Thanks.

Thanks.

Bebop

Can someone show me how to do Question 29 and 30 from Review Exercise 1 of the Heinmann M2 book. I'm following the method my teacher gave me but its coming out wrong. It would be great if someone could show me how to do it, with a good method and the proper way to lay out and approach these questions.

Thanks.

Thanks.

29)a Distance of C.O.M from OA

mass (M) 3m m 2m 4m Total mass = 10m

y-coordinate (y) 0 0 d d/2 y-bar

M x y 0 0 2md 2md 10m(y-bar)

sum of M x y = sum of M x y-bar

4md = 10m x y-bar

y-bar = 4md/10m = 4d/10 = 0.4d

So distance of C.O.M from OA = d - 0.4d = 0.6d

29)b Distance of C.O.M from OC

mass (M) 3m m 2m 4m Total mass = 10m

y-coordinate (y) 0 2d 2d d x-bar

M x y 0 2md 4md 4md 10m(x-bar)

sum of M x y = sum of M x x-bar

10md = 10m x x-bar

x-bar = 10md/10m = d

So distance of C.O.M from OC =d

29)c

Let the angle between the downward vertical and OC be @

tan@ = d/0.6d = 1/0.6 = 5/3

@ = arctan(5/3) = 59 degrees

Sorry it's a bit messy.

for 29 part a).

draw a diagram such as the attachment, so from OA to G is in the y axis.

then you have the ratio of masses (as in the second attachment), and their respective distances from OA.

then take moments- multiply the masses with their respective distances to get:

0.5d*4 + 3*d + 1*d = 10*[y]

[y] = 0.6d

try it for part b).

draw a diagram such as the attachment, so from OA to G is in the y axis.

then you have the ratio of masses (as in the second attachment), and their respective distances from OA.

then take moments- multiply the masses with their respective distances to get:

0.5d*4 + 3*d + 1*d = 10*[y]

[y] = 0.6d

try it for part b).

Fermat

Could you post the question(s) ?

Ok.

29) A uniform rectangular plate OABC has mass 4m, OA = BC = 2d and Oc = AB = d. Particles of mass 2m, m and 3m are attached at A, B and C respectively on the plate. Find, in terms of d, the distance of teh centre of mass of the laded plate:

a) from OA

b) from OC

The corner O of the loaded plate is freely hinged to a fixed point and the plate hangs at rest in equilibrium.

c) Calbulate, to the nearest degree, the angle beteen OC and the downward vertical.

30) The diagram shows an ear-ring made from a unifor swaure lamina ABCD, which has each side of length 4cm. Points X and Y are on the side BC and such that BX = CY = 1cm.

The square portion XYUV is removed and the resulting ear ring is suspended from the corner A. The ear-ring hangs in equilibrium.

The centre of mass of this ear ring is G

a) State the distance, in cm, of G from AB

b) Find the distance, in cm, of G from AD

c) Find, to the nearest degree, the acute angle made by AD with the downward vertical.

Thanks for your help again Widowmaker, I'm finally getting to grips with it. I don't know why I'm finding this stuff so hard, I found P5 a lot easier which i learned in 3 weeks.

Here's 29. I do it like this cos it helps me understand it more in M3.

Sorry couldnt be bothered to resize.

Sorry couldnt be bothered to resize.

Bebop

Ok.

29) A uniform rectangular plate OABC has mass 4m, OA = BC = 2d and Oc = AB = d. Particles of mass 2m, m and 3m are attached at A, B and C respectively on the plate. Find, in terms of d, the distance of teh centre of mass of the laded plate:

a) from OA

b) from OC

The corner O of the loaded plate is freely hinged to a fixed point and the plate hangs at rest in equilibrium.

c) Calbulate, to the nearest degree, the angle beteen OC and the downward vertical.

30) The diagram shows an ear-ring made from a unifor swaure lamina ABCD, which has each side of length 4cm. Points X and Y are on the side BC and such that BX = CY = 1cm.

The square portion XYUV is removed and the resulting ear ring is suspended from the corner A. The ear-ring hangs in equilibrium.

The centre of mass of this ear ring is G

a) State the distance, in cm, of G from AB

b) Find the distance, in cm, of G from AD

c) Find, to the nearest degree, the acute angle made by AD with the downward vertical.

Thanks for your help again Widowmaker, I'm finally getting to grips with it. I don't know why I'm finding this stuff so hard, I found P5 a lot easier which i learned in 3 weeks.

29) A uniform rectangular plate OABC has mass 4m, OA = BC = 2d and Oc = AB = d. Particles of mass 2m, m and 3m are attached at A, B and C respectively on the plate. Find, in terms of d, the distance of teh centre of mass of the laded plate:

a) from OA

b) from OC

The corner O of the loaded plate is freely hinged to a fixed point and the plate hangs at rest in equilibrium.

c) Calbulate, to the nearest degree, the angle beteen OC and the downward vertical.

30) The diagram shows an ear-ring made from a unifor swaure lamina ABCD, which has each side of length 4cm. Points X and Y are on the side BC and such that BX = CY = 1cm.

The square portion XYUV is removed and the resulting ear ring is suspended from the corner A. The ear-ring hangs in equilibrium.

The centre of mass of this ear ring is G

a) State the distance, in cm, of G from AB

b) Find the distance, in cm, of G from AD

c) Find, to the nearest degree, the acute angle made by AD with the downward vertical.

Thanks for your help again Widowmaker, I'm finally getting to grips with it. I don't know why I'm finding this stuff so hard, I found P5 a lot easier which i learned in 3 weeks.

wow, p5 in 3 weeks. good job!

have you gone through past paper centre of mass questions?

Effectively you're just taking moments , if you don't understand the table.

Syncman

Here's 29. I do it like this cos it helps me understand it more in M3.

Sorry couldnt be bothered to resize.

Sorry couldnt be bothered to resize.

Re: 29)c - I think you did arctan(0.6) instead of arctan(1/0.6)

30)

G is the COM of ear-ring

a)

G_x = 2 cm, by symmetry

=========

b)

A = area of square ABCD

a = area of square XYUV

A' = area of ear-ring

Y = dist from AD of COM of A

y = dist from AD of COM of a

Y' = dist from AD of COM of A'

A = 4*4 = 16

Y = 2

M = AY = 16*2 = 32

a = 2*2 = 4

y = 2+1 = 3

m = ay = 4*3 = 12

A' = A - a = 16 - 4 = 12

Balance of moments

A'Y' + ay = AY

12Y' + 12 = 32

12Y' = 20

Y' = 20/12 = 5/3

G_y = 5/3 cm

==========

c)

θ = arctan(G_y/G_x)

θ = arctan(5/3 / 2)

θ = arctan(5/6)

θ = 39.8 deg

==========

G is the COM of ear-ring

a)

G_x = 2 cm, by symmetry

=========

b)

A = area of square ABCD

a = area of square XYUV

A' = area of ear-ring

Y = dist from AD of COM of A

y = dist from AD of COM of a

Y' = dist from AD of COM of A'

A = 4*4 = 16

Y = 2

M = AY = 16*2 = 32

a = 2*2 = 4

y = 2+1 = 3

m = ay = 4*3 = 12

A' = A - a = 16 - 4 = 12

Balance of moments

A'Y' + ay = AY

12Y' + 12 = 32

12Y' = 20

Y' = 20/12 = 5/3

G_y = 5/3 cm

==========

c)

θ = arctan(G_y/G_x)

θ = arctan(5/3 / 2)

θ = arctan(5/6)

θ = 39.8 deg

==========

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