FORCE AND GRAVITY

QUESTION

Solution

Topic 4 Tutorial questions

1. a) Force produced by the wind on the Center of gravity=(

=2880 N

Mass of the wall= () Kg
=1360.8 Kg

Hence weight of the wall= (1360.8) N=13335.84 N

Moment produced by the wind= (2880)/1000

=2592 Nm

Moment produced by the weight across half of the base(in opposite direction) = ()/1000

=600.11 Nm

Since the moment produced by the wind is greater than that produced by the weight hence it is going to fall.

b) If the wind were on the opposite side of the wall there is no change in the likelihood of the wall falling because of the symmetry of the wall the moments would be in the opposite directions to that calculated above but their magnitude same. Hence, the wall would fall in this case too.

2.a) The moment due to the gravity=

The moment due to the Soil force=

Since both the moments are in the opposite direction the retaining of the wall would not be possible.

2. b)The moment due to gravity would change =

2. c) The moment due to the wind =1.3 .

The moment due to the wind would add up with moment caused by the soil.

 

Homework Topic 3 & 4

1. There would be a reaction force of 400 N acting on the Centre of the nut in the upward direction to maintain the force equilibrium.

The moment by the force acting on the arm= () = 112 Nm,

The moment would be lesser in case the force is applied closer from the end.

2.a) The moment equilibrium across the pivot:

() = (3.6)

=4.45 m

 

2. b.) We calculate the force from the moment equation on the either side of the pivot to get P:

() = (0.82)

 

3. The dead load would be equivalent the sum of the weight of the garage and the RCC slabs. This force shall be provided by a pivot as normal reaction between the two weights.

 

4. Similar to above, we have calculated the force from the moment equation on the either side of the pivot:

()=(2.5)

5. Moments into the paper= () +0.4E=30+0.4E

Moments out of the paper= () =6
so, we have E=60 N in the opposite direction.

6. Volume of the concrete block= () =0.648

Mass=0.648*2500 =1620 kg

Suppose the magnitude of the force applied be F, then

 

7.The overturning moment of force (due to earth force) ==0.75KN per unit on the wall

The restraining force (due to the gravity) =

There would no effect on the wall’s position in the current scenario of the applied forces. But if the forces are applied from the side opposite to direction of the earth pressure.

LB40

“The presented piece of writing is a good example how the academic paper should be written. However, the text can’t be used as a part of your own and submitted to your professor – it will be considered as plagiarism.

But you can order it from our service and receive complete high-quality custom paper.  Our service offers Science  essay sample that was written by professional writer. If you like one, you have an opportunity to buy a similar paper. Any of the academic papers will be written from scratch, according to all customers’ specifications, expectations and highest standards.”

Please  Click on the  below links to Chat Now  or fill the Order Form !
order-now-new                               chat-new (1)