Forces question

1. Jun 11, 2007

t_n_p

1. The problem statement, all variables and given/known data

3. The attempt at a solution

Will Fus and Fsu be located on the string in equal and opposite directions?

I can do q (i) with F=ma, but unfortunately I cannot proceed to do so because I cannot answer question h!

2. Jun 11, 2007

Feldoh

Let's think about it. The unicycle rider is towing the surfboard carrier with a string, this creates a tension force. Now, it is the unicycle rider accelerating, and remember that in order for there to be a force two objects have to be in contact (since it's not a non-contact force in this case). Does any of this help?

3. Jun 11, 2007

andrevdh

Yes, a massless string serves to transmit forces from its one one to the other - that is an object pulls on it and it pulls just as hard back on the object. Obviously the string cannot "generate" unequal forces at its ends - it has no internal energy source, it just transmits the force unaltered to its ends. If it does have a weight the attractive force of the earth can alter the tensions at its ends. Also if the string drags over a surface (pulley) with friction the tension at its ends will not be the same since the interaction can change them.

4. Jun 11, 2007

andrevdh

(h) It is "easier" to apply N2 to the system. The tension forces cancels each other out and you therefore do not need to know their values.

5. Jun 11, 2007

t_n_p

Ok, so am I right in saying Fsu and Fus are located "on the string"?

I'm still unsure as to whether to take the system as one.. (lol, andrevdh you got in before I even finished my post!)

Last edited: Jun 11, 2007
6. Jun 11, 2007

t_n_p

Minor double check: when applying Newton's 2nd to find acceleration I only take the horizontal forces, correct?!

I've even forgotten the basics!

Hence I get...

120-10=100a
110=100a
a=11/10m/s^2

7. Jun 11, 2007

andrevdh

That is how I've got it too: 1.1 m/s^2

8. Jun 11, 2007

cool thanks!