2 questions (1 Viewer)

[zxreth]

Question 1:

Two loaded trolleys of masses 3kg and 4kg (which are joined by a light string) are pulled by a spring balance along a smooth horizontal laboratory bench. The reading on the spring balance is 14N

what is the magnitude of the tension in the light string joining the two trolleys?

Question 2:

A warehouse worker applies a force of 420N to push two crates across the floor. The friciton force opposing the motion of the crates is a constant 2N for each kilogram.

Would the worker find it easier to give the crates the same acceleration if the position of the two crates are reversed (note the worker is pushing the crates in the order: 30kg crate first then 40kg crate) ? Support answer with calculations.

Weights of trolleys: heavier one is 40kg
lighter one is 30kg.

Also could you please show all your working out.

Thanks =] much appreciated!

 

[alcalder]

Draw a picture, add all forces and then treat each section of the problem as a separate identity.

Mass 1 Mass 2
| 3kg |--->----<---| 4kg |------------> F_net = 14N

T₁= T₂

Friction = 0N

Look at Mass 2 - because Mass is only a factor in vertical forces

F_net = T₂
=14 N

For question 2 there is no difference in pushing the larger first or the smaller first because the two crates are connected they are essentially just one weight of 70 kg.

Even if the person pushed the end crate and had the second crate lagging behind on a string, the force needed for acceleration would be exactly the same.

Mass 1 = 30kg, Mass 2 = 40kg
Friction 1 = 60N, Friction 2 = 80N
Applied Force = 420N

mass 1 first

F net 1 = Applied force - Friction - Force of 2 on 1
30a = 420 - 60 - F(2 on 1)
30a = 360 - F(2 on 1)
F(2 on 1) = 360 - 3a

Force of 2 on 1 = Force of 1 on 2
Both accelerating at the same rate = a
F(1 on 2) = 360 - 30a

F net 2 = F(1 on 2) - Friction
40a = 360 -30a - 80
70 a = 280
a = 40 m/2²

Now if mass 2 were first

F net 2 = Applied force - Friction - Force of 1 on 2
40a = 420 - 80 - F(1 on 2)
40a = 340 - F(1 on 2)
F(1 on 2) = 340 - 40a

Force of 1 on 2 = Force of 2 on 1
Both accelerating at the same rate = a
F(2 on 1) = 340 - 40a

F net 1 = F(1 on 2) - Friction
30a = 340 - 40a - 60
70 a = 280
a = 40 m/2²

No difference.