How to Find the Reading of a Spring Scale on an Incline
Finding leap balance readings (newtons)
- Thread starter dnt
- Start date
Homework Statement
Find the readings on each of the balances in these four situations (strings are massless and incline is frictionless):
(sorry nearly the photo - dont know how to set up it)
Homework Equations
F=ma, W=mg
The Endeavour at a Solution
1) Seems easy enough. Im pretty sure the balance would just read 98 Due north since the tension above equals the weight of the mass beneath.
2) Im having a little trouble with this one. I cannot tell if its going to be 196 N (98 Northward from each of the weight pulling downwards on each terminate) or just 98 Due north. Each of the blocks has a net force equation of:
F = W - T = ma
98 - T = 0
T = 98 North (T, the tension, is what the calibration will read, right? Thats where the reply comes from?)
so I go the tention in the string is 98 for both but I dont know how to combine the 2 internet force equations into a unmarried respond for the remainder reading.
three) the cake has a net force equation of:
F = W - T - T = ma
Due west - 2T = 0
W = 2T
so the tension in the string/balance reading would exist half of the blocks weight = 49 N.
iv) I know I demand the component of the blocks weight on this one due to the incline. And that should be Wsin30 = 49.
so the cyberspace force equation here is:
F = Wsin30 - T = ma
T = Wsin30 = 49 N
is that the correct reply? What about that other 10 kg block above? How does it get included into the problem (if at all?)
cheers.
Answers and Replies
Regarding (b), realize that the force pulling each side of the spring residue is always the same. It's actually no different than (a).
Regarding (d), that upper ten kg mass is irrelevant, since it'south tied to some support. Over again, the amount of force pulling on each end of the bound residue is the aforementioned: 49 N.
If you're unsure, this is the way I would expect at information technology... examine the freebody diagram of the little claw thing. On one terminate there is the tension in the rope acts on the hook. On the other cease the forcefulness of the jump acts on the hook. The two must be equal and opposite because the hook isn't moving (acceleration is 0... also it is massless... either ane alone is enough for Fnet = 0).
So the tension in the rope attached to the hook is equal in magnitude and reverse in direction to the force the spring exerts... and hence the weight that is read past the scale equals the tension in the rope.
just so i fully understand part (b), how would you evidence the piece of work for it? im still having trouble writing a net forcefulness equation using both blocks (all i did was write it for ane block). it seems unlike than (a) even though its non. ie, why is it non greater since both blocks are pulling on information technology?
im having problem getting part (b) into my caput...this one is confusing and i want to brand sure i empathize it 100%. thanks once again.
And so you simply demand to examine one stop. Information technology doesn't affair which one (every bit long as the system is in equilibrium. in all these situations we have equilibrium).
For the commencement diagram... accept the freebody diagram of the footling claw. What are the forces interim on information technology?
A bound can exist seen equally a symmetrical object. Hooke's police says that a spring exerts a force of F = kx. Just at which end is the force exerted? The reply is both ends... If a leap is horizontal and stretched by an amount x from the unstretched state... then one finish exerts a leftward strength of kx... and the other end exerts a rightward force of kx. This is necessary for the system to be in equilibrium. (if the bound exerts a leftward force of kx... that means it has a rightward force of kx acting on information technology by newton'southward 3rd law... if the leap exerts a rightward forcefulness of kx, then by newton'southward tertiary police force something exerts a leftward force of kx on it... so the two forces on the springs balance).So you lot but need to examine one end. It doesn't matter which one (as long equally the system is in equilibrium. in all these situations nosotros have equilibrium).
For the beginning diagram... have the freebody diagram of the little hook. What are the forces acting on information technology?
just gravity i believe. im starting to get information technology i remember. it just seems that (with the 2d 1) each weight is pulling on the spring so the reading should be double - 100 N pulling on i end, and another 100 North pulling on the other.
just gravity i believe. im starting to get it i think. it only seems that (with the 2nd one) each weight is pulling on the leap then the reading should exist double - 100 North pulling on one terminate, and another 100 Northward pulling on the other.
gravity downwards... and the spring force upward...
the second one isn't different from the first. call back in the first the ceiling is also exerting an upwards strength of ten*g... so even in the first 1 both ends ends are being pulled at...
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