Physics Help Forum Help with physics problems?

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 Aug 16th 2017, 11:01 PM #1 Junior Member   Join Date: Aug 2017 Posts: 3 Force & stress? please help Hi there, I could really use some help to be able to work out some physics problems please! here's an example. i need to be able to see the workings too so that I can work through similar problems please. 1. Q: A child’s swing is connected to a cable with a report breaking stress of 750MNm-2. What mass could be hung from the swing if the diameter of the cable was 8mm? Can you help please? just don't know where to start. Its been 30 years since high school and I just need a hand to get through some of these problems to get back on track. I have a few more that I need help with but just trying to take it one at a time! To work this out we need to understand the force and stress applied? So i Think I need these two formulas but what's next? Force = mass x gravity F = mg, where F is force (Newtons), m = mass (kg) and g = acceleration due to gravity (9.8ms-2) We know the mass so we flip the equation m = 750 x106 / 9.8ms-2 Stress = Force / Area Am I anywhere close? here's another one 2. A box sitting on a shelf is knocked off. The velocity of the box just before it hits the floor is 7.5 m/s. Calculate the height of the shelf using the displacement/velocity/acceleration equation. The package is stationary so its acceleration is zero. Two forces are in play, the force from table pushing upwards (normal force) and gravitational force pushing downwards which are equal so net force is zero. The formula to work out the displacement (distance) is s = ut + ½ at2[1] where s is displacement, u = initial velocity, t = time, and a = acceleration. s = (0 x 7.5m/s) + ½ x acceleration x 7.5m/s2 Any help apprecaited! nd another 3. Propane gas is leaking from a cylinder in the immediate vicinity of an insulated conveyor belt which has built up a charge of 2.5μC. If the conveyor belt has a capacitance of 80pF, is there a danger of the propane being ignited? Consider both electrical energy and voltage in your answer. here;s what i have so far - am I anywhere close? Not even sure how to lay out the answer. have done my best Capacitance Charge can be stored on an object that has capacitance which in this case is 80pF. 80pF = .8 x 10-10 = 0.00000000008F Charge Now Charge (Q) in Coulombs is 2.5μC = 2.5 x 10-6C or 2.5 x 10-8C Voltage V = Q/C where C is the capacitance (in Farads - F) and Q is the Charge (in Coulombs) Since V = Q/C the voltage is: V = 2.5 x 10-8 / .8 x 10-10 = 0.000000025 / 0.00000000008 = 312.5 V Electrical potential energy A charge q which is at a voltage of V volts, has an electrical potential energy W = qV. W = qV = (2.5 x 10-8) x 312.5 = 0.000000025 x 312.5 = 0.0000078125 = 7.81 x10-6 = 7.81mJ Minimum ignition energy in air for acetylene is around 17 mJ . In this case there is little danger of the acetylene being ignited from the electrical potential energy. Voltage risk of ignition/explosion???? Can you please direct me in any way? 4. An alternative to shielding of the radiation source is to extend the distance between the radiation source and the worker. The initial design places the worker 1.2m, from the source. If the design was revised so the worker was 6m from the source, what percentage reduction in radiation exposure will there be for the second position relative to the first? here's what I have - anything right here lol? “To reduce the dose rate by a factor of 4 then the distance to the source must be doubled”[p.18, 3] The distance if multiplied by three reduces the dose rate to 1/32, the distance multiplied by four reduces the dose rate to 1/42 and so on. So at 6m, five times the original distance (5 x 1.2m), the dose rate has reduced to 1/52 or 1/25 of its original value. Distance (m) from source Multiplier Dose rate 1.2 1 Original value 2.4 2 ¼ of original value 3.6 3 1/9 of original value 4.8 4 1/16 of original value 6 5 1/25 of original value 12 10 1/100 of original value The dose rate above is proportional to the inverse square law 1/d2 where d = distance to the source.[3] Position 1: 1/1.22 =69.4% Position 2: 1/62 =2.78% Percentage reduction from position 1 at 1.2m to position 2 at 6m = 66.6%. (is this right?) 5. A 50kg cardboard box needs to be slid across a dusty concrete floor. Joan, a worker can apply a maximum force of 150N. If she is only just able to move the box, what is the coefficient of static friction for the cardboard box on dusty concrete? The next day Joan decides to sweep the floor before trying to move the next box, but she finds that she can’t. What might be the reason? here's what I have so far - anything right here? For Joan to be able to move the box the force applied to the box, F, must be greater than fmax = μ W[4]. W = mg = 50 kg x 9.8 ms-2 = 490 Newtons (N). If CoF [μ] = F/N, where F = horizontal force required to slide a mass across a surface and N = normal (i.e. vertical) force being exerted on the surface by the object, and if Joan can only apply a maximum force of 150N, then the CoF [μ] would need to be a maximum of 0.3 for her to still be able to move the box slightly. F > 0.3 x 490 = 147N Q The next day Joan decides to sweep the floor before trying to move the next box, but she finds that she can’t. What might be the reason? Perhaps the lower frictional resistance available for Joan herself due to less dust on the floor (causing slipping), or in other words the “lateral force applied at the foot-surface is greater than the frictional resistance available”[p.4,4], no longer allows her to apply 150N force or the reduced force now necessary to move the box. ANY HELP APPRECIATED! Last edited by NEWtoPHYSICS; Aug 16th 2017 at 11:41 PM.
 Aug 17th 2017, 03:15 AM #2 Senior Member     Join Date: Jun 2016 Location: England Posts: 618 Your issues seem mostly to do with recognising the correct ways of mathematically manipulating the equations to get the answer you want on the left of the = sign. Q1 You have correctly identified F=ma to get Newtons from Mass You are allowed 750 million Newtons for each square meter of cable cross-section Work out the actual Cross-Section Area of an 8mm diameter cable (in Meters) How much smaller is this actual cross section than One square Meter? This is how much smaller the Force must be than 750 million. This gives you the Maximum Force, use F=ma (where a is the acceleration due to gravity) to find the maximum mass. Q2 Try looking at the problem in reverse. The box starts at 7.5 m/s and decelerates at 9.8m/s/s until stationary. Remember if you differentiate the displacement formula, you get the velocity formula. I will leave other questions for Ron (later Ron) __________________ ~\o/~ Last edited by Woody; Aug 17th 2017 at 09:56 AM.
 Aug 17th 2017, 06:04 AM #3 Senior Member     Join Date: Jun 2016 Location: England Posts: 618 Q4 seems sensible, Q5 part 1 looks OK, part 2 I would suggest that the dust might act (like mini ball bearings) to reduce friction Thus after cleaning the floor is LESS slippy the coefficient of friction has risen it requires more than 150N to move the boxes. Q3 don't know. My practical side would be worried about sparks Not sure if this is what the question is after though. __________________ ~\o/~
 Aug 17th 2017, 04:19 PM #4 Junior Member   Join Date: Aug 2017 Posts: 3 Thanks Woody! I have worked through your comments and have some progress to show. Really appreciate your help. BUT SOMETHING IS REALLY WRONG - WHAT HAVE I DONE WRONG...please help. Question: A safety harness is connected to a cable with a report breaking stress of 800MNm-2. What mass could be hung from a harness if the diameter of the cable was 6mm? We need to use formula Stress = force / area to get the force value. To calculate the area we look to the diameter of cable (6mm), so radius is 3mm. To find area (A) of the cable surface (cross-section) the formula is A = π r2 A = π x 32 = 28.27mm2 (to 2 dp) =28.27 x103m2 (in SI Units of m) We also need to convert the stress into SI units So 800MNm-2 becomes 800 x 106Nm-2 So back to σ = F / A. We know the stress and the area so we need to find force (F). So lets swap the formula around to F = σ A F = σ A = 800 x 106Nm-2 x 28.27 x 10-3m2 = 800 x 106Nm-2 x 0.02827m2 = 22,616,000 N Then to get mass we can use F = ma where F = force, m = m and a = acceleration to rearrange the formula to find mass M = F A = 22,616,000 9.8 = 2,307,755.1kg But surely this is way too big – where have I gone wrong….help please! Last edited by NEWtoPHYSICS; Aug 20th 2017 at 02:53 PM.
 Aug 21st 2017, 10:11 AM #5 Senior Member     Join Date: Jun 2016 Location: England Posts: 618 One mistake I have seen is the number of square millimetres in a square meter You have used 1000 in your calculations, it should be 1,000,000! This will reduce all of your results by a factor of 1000 giving a final answer of 2308kg or about 2 tonnes. I am sure it was just a simple slip of the brain, at least you recognised that there was a mistake. The best way to avoid this sort of thing is (boring) practice, you will eventually build up a pattern of common mistakes and how to find them. __________________ ~\o/~
 Aug 21st 2017, 08:44 PM #6 Junior Member   Join Date: Aug 2017 Posts: 3 Thank you so much Woody!!!! I really appreciate your help! Now it stands out but I couldn't see the wood for the tress before! have a great week.

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