Posterous theme by Cory Watilo

Corrections to 1.23, 1.25 and 1.26 Answers

 

 

1.26 Answer

31 October 2011

17:36

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1.23 Answers PFY p.106 2a, 2b, 3.ppt (589 KB)
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1.26 Answers

1.26 Question

31 October 2011

17:36

·         1.26 understand that the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam 

 

Question 

Image001

·         The Newtonmeters in the diagram labelled L and R are spaced 1m apart 

·         The hanging mass has a weight of 7N and is placed 0.3m from the Newtonmeter labelled R 

·         What are the readings on Newtonmeters L and R? 

 

Hints to help

·         Calculate moments about either of the Newtonmeters; moments should be balanced because the beam is stationary

·         Now calculate moments about the other Newtonmeter

·         You should now have found the readings on each Newtonmeter; is there a way to check that these answers are correct?


 

 

Fa x da = Fc x dc

7 x 0.3 = Fc x 1.0

2.1N= Fc

Fa x da = Fc x dc

7 x 0.7 = Fa x 1.0

4.9N = Fa


The reading on the Newton meter R will be 2.1N

The reading on Newton meter Lwill be 4.9N

 

 

1.25 Answers

 

 1.25 Principle of Moments Questions 

29 September 2010

14:37

 

Size of load (tonnes)

Size of load (N)

Maximum length of jib (m)

Moment (load x length)

10

100,000

12

1,200,000

20

200,000

6

1,200,000

30

300,000

4

1,200,000

40

400,000

3

1,200,000

 

5b) All of the moments are the same, this is because the highest moment the crane can extend to is 1,200,000.

c.  M= F x d, M=1,200,000, d=24m, F= 1,200,000/24, F=50,000N

 

4a) X=20N

b) Y=4N

c) Z=1N

 

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1.23 Answers

1.23 Moments Questions

29 September 2010

14:37

PFY p106

2a) this means that the mechanic will have to apply less force to undo a tight nut, as he is further away from the pivot.

b) This means that a person will have to apply less force to open the door, as the handle is far away from the pivot (hinge)

3. M=F x d, F=200N, d=20cm, M=200 x 20, M=4000Ncm

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1.26

1.26 Question

31 October 2011

17:36

·         1.26 understand that the upward forces on a light beam, supported at its ends, vary with the position of a heavy object placed on the beam

 

1.25

1.25 Starter

31 October 2011

12:12

·         Why doesn't the see-saw balance?

 

Explanation

·         The perpendicular distance from the pivot is the same on each side

·         But the force (weight) on the left is great

·         So the moment (turning force) on the left is greater

·         The see-saw will rotate anticlockwise

 

 

 

1.25

Tuesday, July 06, 2010

3:04 PM

·         1.25 recall and use the principle of moments for a simple system of parallel forces acting in one plane

Principle of Moments

·         When a system is balanced then the anticlockwise moments are equal to the clockwise moments

·         This means that the anticlockwise turning force is equal to the clockwise turning force and the system doesn't move

 

Ma = Mc

Fa x da = Fc x dc

 

Ma = anticlockwise moment (Nm)

Fa = force causing the anticlockwise rotation (N)

da = perpendicular distance of Fa from the pivot (m)

Mc = clockwise moment (Nm)

Fc = force causing the clockwise rotation (N)

dc = perpendicular distance of Fc from the pivot (m)

 

 

 

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