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Can anyone help me with these questions? Take g = 10 ms^-2
1) A trolley of mass 200 kg is being pulled up a smooth slope of 20 degrees by a rope parallel to the slope. If the tension in the rope is 800 N, find the acceleration of the trolley.

(edited 5 years ago)

Reply 1

Z200

OP

11

2) A girl of mass 65 kg is abseiling down a rope fixed to the top of a cliff. If the tension in the rope is 540 N, what is the resultant force on the girl? Find her acceleration. What happens if the rope breaks. 3) A ball of mass 1 kg is falling through the air with an acceleration of 6 ms^-2. Calculate the air resistance. Please can anyone make me understand

Original post by Z200

Can anyone help me with these questions? Take g = 10 ms^-2
1) A trolley of mass 200 kg is being pulled up a smooth slope of 20 degrees by a rope parallel to the slope. If the tension in the rope is 800 N, find the acceleration of the trolley.

Have a go yourself. It's a good idea to first sketch out the scenario and label all the forces and angles.

OP

Original post by Z200

...

Looks good, my only criticisms are

(a) in the diagram you draw the weight straight down because that's direction gravity works in... not in the direction you labelled.

(b) I'm not sure why you said

=10a

. The mass of the trolley isn't 10kg.

(edited 5 years ago)

Reply 5

Z200

OP

11

So it should've been 200 a

(edited 5 years ago)

Original post by Z200

So it should've been 20 a

200a ... The mass is 200kg

OP

OP

2nd question

Original post by Z200

2nd question

Again, your working out is pretty much correct but your diagram does not model the scenario at all. I am not sure where you are pulling the four forces all acting in different directions from!

The person is attached to a ROPE and is descending DOWN... so literally the diagram should be a particle (which models the person) of mass 65kg, which has a force acting straight down that is its weight force, and the tension force acting vertically up. That's it, just two forces.

Having that, you do find that her acceleration is 1.69 ms^-2 indeed. But when the rope breaks, then yes; the net force is entirely

mg

but I think the question wants you to say what the acceleration is now... which is as easy as it sounds without any work necessary.

(edited 5 years ago)

Reply 10

Z200

OP

11

3rd question and would you help with 5 as well please. Thank you.

Original post by Z200

3rd question and would you help with 5 as well please. Thank you.

No, this is wrong. Why do you have the

F_r

force?? What's that supposed to be?

And moreover, how are you solving for this horizontal force

F_r

by using vertical forces...? -- So this makes no sense.

(edited 5 years ago)

OP

Air resistance

OP

I'm not good at doing force diagrams

Original post by Z200

Air resistance

So why is it acting horizontally? And so what is

F

??

Air resistance opposes motion, so it acts in the opposite direction from where the ball is going.

This is the same case as with the cliff climbing person, there should only be two forces.

OP

Original post by Z200

...

Yep, so now work out the correct

F_r

value.

OP

so Fr = 4 N

Original post by Z200

so Fr = 4 N

Yep, easy as that.

For Q6, in the case when the lift is moving with constant speed, what is the acceleration at which the woman is going up?? Can you model the forces on her body?

Hence answer the questions.

Reply 19

Z200

OP

11

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