Solving Basic Tension Problems: Newton Readings on Calibrated Spring Balances

[maniacp08]

In the figure below, the objects are attached to spring balances calibrated in Newtons. Give the readings of the balances in each case, assuming that the strings are massless and the incline is frictionless.

Im kinda confuse on what the question ask but here is what I think.
Does this relate to Hooke's Law Fx = -KX?

If it is there is 2 unknowns. I know Fx = to the weights but X?

Could you tell me if this is correct?
For
A) is it 10N?
B) 0N
C)5N
D) ?

 

[edziura]

"Does this relate to Hooke's Law Fx = -KX?"

Not really. The spring balance in each case measures the tension in the strings. In each case there is mechanical equilibrium. The question is really: How much tension is in the string if the masses in each situation are in mechanical equilibrium?

One other point: don't forget the relationship between mass (kg) and weight or force (N).

 

[maniacp08]

How much tension is in the string if the masses in each situation are in mechanical equilibrium?

Could you elaborate on that? I am still confuse on what you mean by that

 

[Redbelly98]

You'll need to draw a force diagram for each mass.

The masses are not moving since they are in equilibrium. That simplifies the well-known equation F=ma.

The string tension will enter into the net force. Since the string tension is the balance reading, solving for the tension will answer the questions.

Your answers are incorrect, however you are pretty close on (a) and (c). Question: what is the weight of a 10 kg mass?

 

[maniacp08]

10kg * 9.81 for A
and for C is the answer for A/2?

F = MA
F = M * 9.81?

How come for B is not 0? and How would I approach for D?

 

[Redbelly98]

10kg * 9.81 for A
and for C is the answer for A/2?

Yes, pretty much, except that it is 9.81 N/kg, not just 9.81 (if you want to do the units properly, that is).

F = MA
F = M * 9.81?

The acceleration of an object at rest is not 9.81.
Draw the force diagram for the masses.

Net Force = MA, and A = ____ for an object at rest.

How come for B is not 0? and How would I approach for D?

B: the string tension is not zero. Draw a force diagram for each mass.
D: Draw a force diagram for each mass.

 

[maniacp08]

For D.
The force of gravity is 98.1N
The Normal force is cos(30) * 98.1N

Since there is 10kg mass pulling on the string, I need to find the force
so is sin(30) * 98.1N which is 49.05N

Is that the answer?

 

[Redbelly98]

For D.
... 49.05N

Is that the answer?

Yes.

 

[maniacp08]

Thanks so much for helping! =]