Forces on an Accelerating Object

[sweet877]

A 2.1 X 10^-4 kg spider is suspended from a thin strand of spider web. The greatest tension the strand can withstand without breaking is 2.0 X 10^-3 N. What is the maximum acceleration with which the spider can safely climb up the strand?

Fnet = ma = (2.1 X 10^-4)(a) < 2.0 X 10^-3
a < 9.52 m/s^2?

 

[PhY_InTelLecT]

Is the spider hanging vertically? Is there no gravity? If there is, then the spider can't even climb up the web. As you can see, gravity alr require a greater acceleration than the ans that u have found. If there is gravity, the web will break under the spider's own weight...

 

[sweet877]

There is gravity...hmm...
So the force the spider is exerting w/ 0 acceleration is mg = 2.1 X 10^-4 (9.8) = 2.06 X 10^-3
Fnet = ma = (2.1 X 10^-4)(a)

 

[PhY_InTelLecT]

ya.. The Weight of the spider as calculated by u, 2.06 X 10^-3N, is already more than the web can actually support.. Unless the qn requires you to assume that gravity is not present, then your previous solution is right..

 

[sweet877]

I see...thanks for your help!

 

[PhY_InTelLecT]

If gravity is present.. Your working muz be this
Fnet= mg + ma < 2.0 X 10^-3..
By taking g as 9.81ms^2, you will see that , a will be negative. which means that the spider must decelerate..ya?

 

[PhY_InTelLecT]

oh, no problem at all.

 

[sweet877]

Wait...wouldn't gravity be negative though?

 

[PhY_InTelLecT]

The -ve sign of gravity actually defines the direction of the force in which it is acting. However, when you encounter qns like this, it all comes down only to the absolute value, magnitude. Since both the g and a are in the same direction, the -ve sign does not matters anymore.
What you will get is just a negative value for Fnet, but note: the negative doesn't represent the value!, It just shows that the Fnet that u found is at an opposite direction to tension. Tension is upwards, thus +ve, while Fnet is downwards, thus -ve. And when T-Fnet>0 in order not for web to break, T > Fnet.

 

[sweet877]

Oh OK...I get it now. Thanks!

 

[PhY_InTelLecT]

glad u got it.. Haha:)