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### 02week02

Course: CIV 1501, Fall 2009
School: Allan Hancock College
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Word Count: 669

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ENGINEERING CIV1501 STATICS Week 2 Lecture Content Semester 2, 2002 NOTE The first question in the final exam shall be taken from this topic. 1. Force Force is something that tends to cause a mass to accelerate can be push (compression) or pull (tension). Defined as F = ma Weight is the force caused by the acceleration due to gravity. Thus W = m g , where g is force due to gravity = 9.82 m/s2 Hence the weight...

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ENGINEERING CIV1501 STATICS Week 2 Lecture Content Semester 2, 2002 NOTE The first question in the final exam shall be taken from this topic. 1. Force Force is something that tends to cause a mass to accelerate can be push (compression) or pull (tension). Defined as F = ma Weight is the force caused by the acceleration due to gravity. Thus W = m g , where g is force due to gravity = 9.82 m/s2 Hence the weight of 1 kg = I kg x 9.8 m/s2 = 9.8 N It may be helpful to think of a newton as about the weight of 100 gm, i.e. the weight of a small mobile phone or calculator, or of a small Jonathon apple (a falling apple is supposed to have given Newton the idea of gravitational acceleration). REMEMBER if force or weight acts on a body, the body will accelerate or move unless the force is cancelled out by an equal and opposite force. Thus if I push on a wall with a force of 40 N, the wall must push back with exactly that, or else it will fall over (or I will fall backwards). If a force or weight acts and is not cancelled out, the body must move. Example - A car with a mass of 1.3 tonnes is towed by a truck on level ground. If the rope tension is 2.45 kN and the drag force (caused by friction and wind resistence) is known to be 550 N, what is the acceleration of the towed vehicle? Solution - 1.3 tonnes = 1300 kg and 2.45 kN = 2450 N Unbalanced force = 2450 550 = 1900 N F 1900 F=ma = a = = 1.46 ms-2 (or about half the m 1300 acceleration of a normal car, i.e. 0 to 100 kph in 10 secs gives 2.8 m/s2 acc) 2. Resultant of Concurrent Forces Example 1 - (2-dim) An advertising sign is anchored by three cables to a single point on a nearby wall. Find the magnitude and direction of the single force that is equivalent to the three forces acting on the wall anchor. The diagram is with the solution. Solution to Example 1 Using the parallelogram rule, could one combine forces A and B to obtain a resultant, then combine this resultant with C to get the final resultant. It probably would be a good idea to also do this graphically by hand to get some idea of the sort of solution you should obtain. However, the best way to combine more than two forces in any 2-D example is to use vectors as below. This is the method preferred in this course, although other methods are permitted. Example 2 - (3-dim) Two cables pull on the corner A of a building as shown in the diagram with the solution. What is the magnitude and direction of nthe single resultant if the forces in the cables are AB = 33 kN and AC = 28 kN? Solution to Example 2 Procedure 1. Use distances to get the unit vectors or direction vectors 2. Multiply the unit vector by the magnitude of the applicable force to obtain the force components. 3. Add the force components to get the resultant in vector form 4. Combine the components to obtain the magnitude and direction of R. 3. Equilibrium of Concurrent Forces A particle is in equilibrium when all the forces acting upon it cancel each other out. R...

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