Balancing Forces on a String: Can Fus and Fsu be Equal and Opposite?

[t_n_p]

Homework Statement

http://img263.imageshack.us/img263/1947/untitledtq5.jpg

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!

 

[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?

 

[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.

 

[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.

 

[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!)

 

[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

 

[andrevdh]

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

 

[t_n_p]

cool thanks!