Tension? (2 Viewers)

[NubMuncher]

A little confused on how to work out tension

Say someone is carrying a weight of 1.5 kg upwards using a piece of string. He is accelerating upwards at .5 m/s/s.

My Working
SumF = m x a
=0.75 N upwards

After this I am led to believe that you must add that value to the bricks weight, i.e.
0.75 + (-14.7) = 13.95 N downwards

Any help appreciated

 

[MrBrightside]

ahaha when i first saw this thread i thought it was about exam tensions and study pressure lol If only that could be worked out as with the qs sorry bud, maybe someone else will reply soon. : )

 

[ohexploitable]

edit: hahaha disregard that i suck cocks

edit 2: or maybe not...

edit 3: actually, i do suck cocks, my solution is long and stupid, still got the right answer though...

 

[cutemouse]

Say someone is carrying a weight of 1.5 kg upwards using a piece of string. He is accelerating upwards at .5 m/s/s.

Draw a diagram showing the forces on the mass (ie. weight, mg and Tension, T) and indicate the positive direction.

If upwards is positive then:

0.5 * 1.5 = T - mg

T = 0.5*1.5+1.5*9.8 = 15.45 N upwards

BTW this is in the preliminary course.

 

[FCB]

f=mg+ma
1.5(0.5+9.8)+15.45N

 

[FCB]

hahaha disregard that i suck cocks

Im pretty sure that was correct.

Edit, i mean your answer before you edited your post. Not the sucking cock bit

 

[NubMuncher]

Isn't gravity in the negative direction? Therefore:
T = .75 + (-14.7)
T = -13.95 upwards
T = 13.95 downwards
?

 

[ohexploitable]

Im pretty sure that was correct.

I saw jm01's solution and he's in uni so I automatically assumed I was wrong...

 

[Bored Of Fail 2]

I saw jm01's solution and he's in uni so I automatically assumed I was wrong...

unfortunately he is right lol

 

[ohexploitable]

Isn't gravity in the negative direction? Therefore:
T = .75 + (-14.7)
T = -13.95 upwards
T = 13.95 downwards
?

Think about this, you're holding a mass attached to a rubber band, you suddenly pull the rubber band upwards, it will stretch right? Therefore, when you are pulling the rubber band (accelerating it upwards) there is a greater force on the rubber band (more tension).

 

[FCB]

Isn't gravity in the negative direction? Therefore:
T = .75 + (-14.7)
T = -13.95 upwards
T = 13.95 downwards
?

The mass is accelerating upwards with net acceleration, a. Here, the tension upwards in the string is doing two jobs. It is fully balancing the weight force downwards and supplying the required force upwards to accelerate the mass with the net acceleration of a.


Clearly, with the mass accelerating upwards, the tension, T = m ( g + a ).

 

[ohexploitable]

unfortunately he is right lol

No, he edited his message, he originally said like 12 N.

 

[cutemouse]

I saw jm01's solution and he's in uni so I automatically assumed I was wrong...

Sorry I made an arithmetic error. Serves me right for doing it in my head

 

[jyu]

A long rope is pulled with a force of 20 N at one end and 15 N at the other end. Is this possible and what can you say about the tension in the rope?

 

[cutemouse]

A long rope is pulled with a force of 20 N at one end and 15 N at the other end. Is this possible and what can you say about the tension in the rope?

Yes it is possible. Tension in the rope is the same throughout the rope (provided that it's inextensible).

 

[jyu]

so what is the tension?

 

[Fizzy_Cyst]

A long rope is pulled with a force of 20 N at one end and 15 N at the other end. Is this possible and what can you say about the tension in the rope?

Whilst it is possible, you really would need to attribute certain conditions to the long rope in order to get a 'proper' answer.

In Stage 6 Physics, we assume that ropes/strings and the like are ideal, i.e., always massless, taut and not encountering any resistive forces, and hence the physics of these systems can be approximated.

The answer that you would probably be looking for is 15N Tension Force.

Considering the rope as 'the system', say 20N is acting to the left and 15N to the right. The net force on the system is 5N Left. Seeing as the net force is 5N left and there is an applied force of 20N Left, the Tension force (assuming there is no friction or other resistive forces), must be 20-5N Right = 15N

However, if the net force was 5N Left and we are considering the rope to be ideal, i.e., massless, then think about this in terms of N2L.. doesn't quite make sense

But, then if we do attribute a mass to the rope and the distribution of the mass within the rope, the answer may be a little different, but dont worry about this until 1st year university

 

[jyu]

Whilst it is possible, you really would need to attribute certain conditions to the long rope in order to get a 'proper' answer.

In Stage 6 Physics, we assume that ropes/strings and the like are ideal, i.e., always massless, taut and not encountering any resistive forces, and hence the physics of these systems can be approximated.

The answer that you would probably be looking for is 15N Tension Force.

Considering the rope as 'the system', say 20N is acting to the left and 15N to the right. The net force on the system is 5N Left. Seeing as the net force is 5N left and there is an applied force of 20N Left, the Tension force (assuming there is no friction or other resistive forces), must be 20-5N Right = 15N Right. (N3L - considering the LHS and RHS of the rope as seperate objects)

However, if the net force was 5N Left and we are considering the rope to be ideal, i.e., massless, then think about this in terms of N2L.. doesn't quite make sense

But, then if we do attribute a mass to the rope and the distribution of the mass within the rope, the answer may be a little different, but dont worry about this until 1st year university

Ok
Let say the rope has a uniform density and it lies on a frictionless horizontal surface. These were the unspecified assumptions when the question was posted.

15 N is not the answer. It can be answered using Newton's laws.

 

[jyu]

Not 30 N.

 

[Fizzy_Cyst]

Ok
Let say the rope has a uniform density and it lies on a frictionless horizontal surface. These were the unspecified assumptions when the question was posted.

15 N is not the answer. It can be answered using Newton's laws.

I fail to see how the maximum tension could not be 15N? If the wire is accelerating, the tension in the wire will actually vary throughout the wire, even if of uniform density :\ iirc

Maximum tension should be the same magnitude as the lowest reaction force, which in this question is 15N. Please dont say the answer is 20N due to T2-T1 = ma..

Anyway, this question would be beyond what is required in the syllabus as they do not look at wires which are not in equilibrium