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November 01, 2025, 09:26:37 am
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atom

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Help needed
« on: April 28, 2012, 10:58:54 pm »
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I need help with this question:

A boat of mass 1000kg tows a dingy of mass 250kg.  A thrust force of 5400N is provided by the boat’s motor.  The drag force on the boat is 900N and that on the dingy 300N.  Assume the air resistance for both bodies are negligible. Find the force with which the dingy is pulled by the boat (ie. the tension in the rope.)

Thanks in advance :)

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Bhootnike

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Re: Help needed
« Reply #1 on: April 28, 2012, 11:05:52 pm »
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Quote from: atom on April 28, 2012, 10:58:54 pm
I need help with this question:

A boat of mass 1000kg tows a dingy of mass 250kg.  A thrust force of 5400N is provided by the boat’s motor.  The drag force on the boat is 900N and that on the dingy 300N.  Assume the air resistance for both bodies are negligible. Find the force with which the dingy is pulled by the boat (ie. the tension in the rope.)

Thanks in advance :)

Any answer given by any chance?  I did it in my head cause I'm on my phone but yeah i got 1380.  Not sure if it is right though
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atom

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Re: Help needed
« Reply #2 on: April 28, 2012, 11:16:16 pm »
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Sorry but I don't have the answer to the question, but can you please show me how you got your answer? Thanks.
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DisaFear

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Re: Help needed
« Reply #3 on: April 28, 2012, 11:20:50 pm »
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The Net Force is 5400N - 900N - 300N = 4200N to the right
The two masses are connected, so they have the same acceleration
Net Force = mass x acceleration
4200N = (1000kg + 250kg) x a
a = 3.36 m/s^2

The acceleration of  the smaller boat is 3.36m/s^2. It has a mass of 250kg. What is the Net Force on it?

Net Force = 250kg x 3.36 m/s^2 = 840N
This is the NET FORCE on the object. Not the tension. Remember, we have 300N of drag opposing the motion

The forces opposing motion and the forces helping motion should add to 840N

840N = T - 300N
T = 1140N


Curious, how'd you get that answer Bhootnike? I could be wrong
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Bhootnike

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Re: Help needed
« Reply #4 on: April 28, 2012, 11:29:59 pm »
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I did the exact same as you but I didn't take away drag forces at the start.
So a=5400 / 1250 = 4.32 m/s^2
F=4. 32 x 250
= 1080
+  300

=1380

But I think you're right!
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DisaFear

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Re: Help needed
« Reply #5 on: April 29, 2012, 10:42:44 am »
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Keep in mind, the formula F=ma is for net forces. You'll have to resolve the forces before you can go about using it. Something I had trouble with too in VCE
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Bhootnike

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Re: Help needed
« Reply #6 on: April 29, 2012, 05:53:06 pm »
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oo ok, well yeah, good i looked here and got it wrong!
:D
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khuda ne jab tujhe banaya hoga, ek suroor uske dil mein aaya hoga, socha hoga kya doonga tohfe mein tujhe.... tab ja ke usne mujhe banaya hoga
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