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Spring scale- confusing me... help |
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| Aug26-06, 08:48 AM | #1 |
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Spring scale- confusing me... help
Hi
firstly Thanks for the help..... I am not very good at forces ... and spring balances etc.... Q1 A 10N force is pulling up on the ring of spring scale that weighs 2N. If an 8N mass is attached to the bottom hook of the scale, what is the scale reading? I figure that the mass of the spring is not relevent - hence the answer should be 10+8 = 18N. But the answer is 8N ! Hence I can't solve the next one also: Similar as Q1 above but this time ithe force pulling up the ring of the spring scale is 5N instead.... I am really confuse about these. please help.... thanks
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| Aug26-06, 09:05 AM | #2 |
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| Aug26-06, 06:05 PM | #3 |
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The first bit is simpler than it sounds, matthew, maybe it's rather a trick question. Imagine you're holding up the scale, using a force of 2N because that's what it "weighs". Now you hang another 8N on the hook, and you're now exerting 8+2=10N. But the scale says 8N because that's whats on the hook.
Actually, the second bit is easier than it sounds too. I won't tell you the answer. But here's a clue: equal and opposite. |
| Sep2-06, 09:00 AM | #4 |
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Spring scale- confusing me... help
Ok, let me try
Q1 - there is an upward force of 10N. there are two downward forces - the spring 2N and the stone 8N. So Nett force is zero. Hence the spring reading is the reading of the stone which is 8N. am I right in my understanding? Q2. Upward force = 8N, Downward force = 2N (mass spring) + 8N (stone) Nett force 10-8N = 2 N downward so the spring reading is 2N Somehow, it doesn't feel right. Please help me to correct.... cheers and thanks |
| Sep3-06, 07:37 AM | #5 |
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Use that to deduce the force that the scale must exert on the hanging mass. Which is also the force that the hanging mass exerts on the scale--via Newton's 3rd law. And that's the force that the scale will read. |
| Sep3-06, 05:12 PM | #6 |
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Try this again:
Q2. Upward force = 5N, Mass spring = 2N, Mass stone = 8N Downward force = 2N (mass spring) + 8N (stone) = 10N Nett force 10-5N = 5 N downward so the spring reading is 5N Is this correct? Thanks a lot. Cheers! |
| Sep4-06, 06:33 AM | #7 |
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Do exactly as I said. First find the acceleration of the system using Newton's 2nd law. Hint: You know the weight of the scale and the hanging mass--but what are their masses? |
| Sep4-06, 09:36 AM | #8 |
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Try.....
weight of scale = 2N => mass = 0.2kg weight of stone = 8N => mass = 0.8 kg total mass = 1.0kg now what.... a = F/m which force do I use to find a? 5 N upwards pulling force ? a = 5 m/s2 upwards or use nett force 5N down a= 5 m/s2 downswards either way .... what do I do next? thanks for defusing my confusion...... |
| Sep4-06, 12:56 PM | #9 |
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To find the acceleration of the scale + stone, use the net force acting on it. Once you have the acceleration, analyze the forces acting on the stone. (Hint: There are two forces acting on the stone.)
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| Sep4-06, 06:16 PM | #10 |
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OK,
using the nett force 5N downwards , a = 5 m/s2 down. the stone has an upward force of 5N and its own weight 8N down. Nett force on stone is F = 8-5 = 3N down. So the mass reaading on scale is F/a = 3/5 kg?? Am I getting anywhere ??? |
| Sep4-06, 07:48 PM | #11 |
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Since you know the stone's mass and acceleration, what must be the net force on it? |
| Sep7-06, 10:47 PM | #12 |
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Mass of stone = 0.8kg,
Force Nett = 5 m/s2 down. So the reading on the scale = F = ma = 0.8 x 5 = 4 N. Am I right? Cheers! |
| Sep8-06, 06:58 AM | #13 |
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The scale doesn't read the net force, it reads the force that it exerts on the stone. Let's figure it out. What forces act on the stone? There are two: (1) The scale force acting upward, which is what we are trying to find. Let's call that F_s. (2) The weight acting downward, call it F_w, which equals 8 N. Calling up to be positive, what's the net force in terms of these two forces? F_{net} = F_s - F_w But we already calculated F_{net} above to be -4 N. You do the last step: plug in the values and solve for F_s. |
| Sep9-06, 08:42 AM | #14 |
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Thanks. Trying.....
Force on hook of scale = 5N up Weight of scale = 2 N down ===> mass of scale = 0.2kg Weight of stone = 8N down ====> mass of stone = 0.8kg Force nett of above = 10N-5N = 5 N down. Mass of scale and stone = 1 kg Hence Nett acceleration of system = F/m = 5 m/s2 down. Hence Force nett on stone F(nett) = ma = 0.8kg x 5 m/s2 = 4N down. If up is +ve, then F(net) = -4 N F (on weight if stone) = - 8N (given) Forces On stone: F(nett) = F (scale pulling stone up) - F (force on stone mass) -4 = F (scale) - (-8) F (scale) = -4 - 8 = -12 N. Hence reading from scale = -12N How am I doing now ???? |
| Sep9-06, 09:33 AM | #15 |
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F(net) = F (scale; up) + F (weight; down) -4 = F (scale) + (-8) = F (scale) - 8 Almost there.
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| Sep9-06, 09:54 AM | #16 |
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But earlier it was F(net) = F(scale; up) - F(weight; down) why "-" is changed to "+" now? thanks. PS: can u post a similar Q so that I can test if I really understand this.... Cheers!
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