Conneting Particles- Vertically- Help!

Could someone please do Questions 3 and 6 from Ex 5C in the edexcel heinemann book. They are both connecting particles. I can do horizontal objects, but not when they are vertically connected. Could you please explain as you go?

Lots of rep available for nice people! hehe

Thanks everyone!

Reply 1

SsEe

The only difference is that you have to consider the component of weight on each particle.
If you're still stuck then you'll have to post the questions as I don't have a copy of the book.

Reply 2

faa

3) A particle A of mass 0.5 kg is suspended by a vertical string. A particle B of mass 0.4 kg is suspended from A by means of another string. A force of 10N is applied to the upper string and the particles move upwards. Find the tension in the lower string and the accelaration of the system.

6) Block A of mass 100kg is suspended by a vertical cable. A block B of mass 150 kg is suspended from A by means of a second vertical cable. The blocks are raised 10 m in 10 seconds, starting from rest. Find the tension in each cable.

Reply 3

SsEe

3)
^ 10N
A
V 0.5g
V T
|
^ T
B
V 0.4g

All motion is upwards. So:
For A:
F = ma
10 - 0.5g - T = 0.5a
For B:
T - 0.4g = 0.4a

We have simultaneous equations. Adding we get:
10 - 0.9g = 0.9a
a = (100-9g)/9 = 59/45 = 1.31 ms^-2
Subbing back into the equation for B:
T - 0.4g = 0.4(59/45)
T = 40/9 = 4.44 N

Reply 4

SsEe

The blocks are raised 10m in 10 seconds. ALl forces are constant so acceleration is constant. So we can use the suvat fomulae to calculate the acceleration of the system:
s - 10
u - 0
v
a - ?
t - 10
s = ut + 0.5at²
10 = 50a
a = 0.2 ms^-2

Considering only B:
F = ma
T - 150g = 150a = 150*0.2 = 30
T = 30 + 150g = 1500 N

Call the tension in the upper cable K:

^ K
A
V 100g
V T
|
^ T
B
V 150g

Now consider A
F = ma
K - 100g - T = 100a
K - 100g - 1500 = 100*0.2
K = 2500 N