# Homework Help: Obstacle course acceleration problem

1. Jan 11, 2010

### Helpme7534

1. The problem statement, all variables and given/known data
While running an obstacle course, Jane comes to a rope hanging from a platform. She grabs a rope, which then accelerates her up into the air. Her mass is 55 kg and she accelerates upward at 1.50 m/s2. The tension on the rope is ____N.

2. Relevant equations
F=ma

3. The attempt at a solution
F=55(1.50)
F=82.5 N

Is this right?

2. Jan 11, 2010

### jegues

Re: Acceleration

No, when applying Newtons 2nd law you need to consider the sum of the forces acting on a body.

So,

$$\sum F = ma$$

3. Jan 11, 2010

### Helpme7534

Re: Acceleration

can you show me how to do a question like this?
(keep in mind i only have physics 20-1)

4. Jan 11, 2010

### jegues

Re: Acceleration

What forces are acting on the object when she grabs the rope?

Let me remind that we are given her acceleration in the upwards direction so we really only need to consider the vertical forces acting on the body.

5. Jan 11, 2010

### Helpme7534

Re: Acceleration

the force of her and the force of gravity?

6. Jan 11, 2010

### jegues

Re: Acceleration

The force of tension and the force of gravity. The force of gravity can also be denoted as her weight.

Now apply newtons 2nd law and solve for the force of tension.

7. Jan 11, 2010

### Helpme7534

Re: Acceleration

F=539.55N
is that the ans?

8. Jan 11, 2010

### jegues

Re: Acceleration

No.

How about instead of posting one line of "Is this the answer?" you show us your work in full detail and your reasoning behind it.

Only then will we be able to truly help you.

(By the way, I don't consider handing you the answer helping)

9. Jan 11, 2010

### Helpme7534

Re: Acceleration

dude i did what u told me to do i wasnt randomly posting that
but if u cant see that..

F=ma
F=her mass (aka 55kg, since you may not wan to scroll to the top)(9.81m/s2)
F=539.55 N

(yes im grouchy but its only cus ive only been through this with other ppl who after a waste of my time can't figure it out themselves in the end, i apologize)

10. Jan 11, 2010

### jegues

Re: Acceleration

$$\sum F = ma$$

It's the SUM OF THE FORCES on the body that will correspond to its mass times its acceleration.

You told me there were two forces acting on her body (atleast in the vertical direction).

If you defined the y axis to be positive upwards and negative downwards, sum the forces on the left handside of the equation according to their direction and on the right hand side simply compute her mass times her given acceleration.(keeping in mind the way you've defined your y axis)

11. Jan 12, 2010

### Helpme7534

Re: Acceleration

so

(1.50)(55)-(9.81)(55)=-524.835N

12. Jan 12, 2010

### jegues

Re: Acceleration

$$T - F_{g} = ma$$,

You need to review/practice your application of Newton's 2nd law. Are you drawing free body diagrams when you're attempting a problem?

13. Jan 12, 2010

### Helpme7534

Re: Acceleration

So T=622.05

i didnt get that from no where i went T=ma+Fg

14. Jan 12, 2010

### jegues

Re: Acceleration

$$F_{g} = mg$$

15. Jan 12, 2010

### Helpme7534

Re: Acceleration

so T=(55)(1.50)+(55)(9.81)

16. Jan 12, 2010

### jegues

Re: Acceleration

Correct.

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