Some M1 - statics of a particle questions

Hi can someone please help me with the following questions - rep will be give and thanks in advance.

6) Newtons cradle consists usually of five small steel spheres each suspended by two light strings. Modelling this by a number of particles each suspended by two strings inclined at 80 degrees to the horizontal, find the tension in each string when the mass of each steel sphere is 0.1 kg.

7) A peg bag of mass m hanging from a washing line may be modelled by a particle supported in equilibrium by two straight strings inclined at alpha and beta to the horizontal as show in the diagram. find the tension in each string when :

m = 0.5
alpha = 10 degrees
beta = 12 degrees

diagram attached

I need to know how and why the answer are what they are - find this tension thing quite hard.

Regards, Asad

untitled.bmp73.0KB

Reply 1

mizfissy815

12

What past paper/ book question?

Reply 2

asadtamimi

OP

13

mizfissy815

What past paper/ book question?

Ex4D p.84 Q's 6b + 7b on heinemann m1 modular maths edexcel book

Reply 3

mizfissy815

12

6a) resolve horizontally
T₁cosθ -T₂cosθ= 0
T₁cosθ =T₂cosθ
so T₁ = T₂

resolve vertically-
2Tsinθ= 0.8
T= (0.8)/(2sin80)
T=0.406

b) 2Tsin80= 0.1 x 9.81
T= (0.981)/2sin80
T=0.498

7) b) resolve horizontally
T₁cosθ -T₂cosØ= 0
T₁cosθ =T₂cosØ
T₁cos10 =T₂cos12
T₁ = (T₂cos12)/ cos10

resolve vertically-

T₁sin10 + T₂sin12=0.5 x 9.81
T₁sin10 = 4.905 - T₂sin12
T₁ = (4.905 - T₂sin12)/ sin10
So...
(T₂cos12)/ (cos10) = (4.905 - T₂sin12)/ (sin10)
T₂sin10cos12 = (4.905 - T₂sin12) x cos10
T₂sin10cos12 = 4.8305 - T₂sin12cos10
T₂sin10cos12 + T₂sin12cos10= 4.8305
0.374606593 T₂ = 4.8305
T₂= 12.89 N≈ 12.9N

T₁ = (4.905 - T₂sin12)/ sin10
T₁ = (4.905 - 12.89sin12)/ sin10
T₁ = 12.8 N

Sorry it took a while...I got side-tracked.

Reply 4

asadtamimi

OP

13

mizfissy815

6a) resolve horizontally
T₁cosθ -T₂cosθ= 0
T₁cosθ =T₂cosθ
so T₁ = T₂

resolve vertically-
2Tsinθ= 0.8
T= (0.8)/(2sin80)
T=0.406

b) 2Tsin80= 0.1 x 9.81
T= (0.981)/2sin80
T=0.498

7) b) resolve horizontally
T₁cosθ -T₂cosØ= 0
T₁cosθ =T₂cosØ
T₁cos10 =T₂cos12
T₁ = (T₂cos12)/ cos10

resolve vertically-

T₁sin10 + T₂sin12=0.5 x 9.81
T₁sin10 = 4.905 - T₂sin12
T₁ = (4.905 - T₂sin12)/ sin10
So...
(T₂cos12)/ (cos10) = (4.905 - T₂sin12)/ (sin10)
T₂sin10cos12 = (4.905 - T₂sin12) x cos10
T₂sin10cos12 = 4.8305 - T₂sin12cos10
T₂sin10cos12 + T₂sin12cos10= 4.8305
0.374606593 T₂ = 4.8305
T₂= 12.89 N≈ 12.9N

T₁ = (4.905 - T₂sin12)/ sin10
T₁ = (4.905 - 12.89sin12)/ sin10
T₁ = 12.8 N

Sorry it took a while...I got side-tracked.

thank you so much - cant give you rep at the moment, says i have to spread it around a bit lol but help greatly appreciated