Chapter 2.9 Equilibrium Presentation

7/31/2019 Chapter 2.9 Equilibrium Presentation

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the Newtons first law body remain at restwhen a net force of zero on it. When net forceis zero, the forces are in balanced or


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The force of gravity is acting on it and thereis another force (equal in magnitude andopposite in direction) acting to balance theweight, so the net force is zero.


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why net force is zero when a car moving withconstant speed and a wall pushed


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a football player kicks a ball horizontally witha force of 140N towards the east. The ballwas located on thick mud so that a frictionalforce of 30N towards the opposite direction.

Then teacher asks what the net force on it?


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Diagram above shows that four forces ofmagnitude 2N, 4N, 5N and 8N are acting onpoint O. All the forces are perpendicular toeach others. What is the magnitude of theresulatant force that acts on point O?


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example when there are two forces areperpendicular (or any other angle) to eachother, the net force can be found byperforming a vector addition.

80N to East, 60N to South.


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80N

60N


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80N to East, 60N to South.

Angle between South and East = 37

Net Force (Southeast) = 100N


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Fnet = Fx + Fy


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1 . Fnet = Fx + Fy 2. Fx = 5N-2N = 3N

3. Fy = 4N-8N = -4N(-4N downwards)

. Fnet = 5N ( to South east)


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Diagram above shows a lorry pulling a logwith an iron cable. If the tension of the cableis 3000N and the friction between the log andthe ground is 500N, find the horizontal force

that acting on the log.

HCos = A Fx = 3000Cos30 500


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Horizontal component of the tension = 3000cos30o =2598NFriction = 500N

Resultant horizontal force = 2598N - 500N=2098N

Fx = 2098N (to right)

Fnet = 2098 0 = 2098N (to right)


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First of all, let's examine all the forces orcomponent of forces acting along the plane.

The force pulling the block, F = 20NThe frictional force Ffric = 2N

The weight component along the plane =20sin30o = 10N

The resultant force along the plane = 20 - 2- 10 = 8N


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Diagram above shows two forces ofmagnitude 25N are acting on an object ofmass 2kg. Find the acceleration of object P, inms-2.


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Horizontal component of the forces =25cos45o + 25cos45o = 35.36N

Vertical component of the forces =25sin45o - 25sin45o = 0N

The acceleration of the object can bedetermined by the equation

F = ma

(35.36) = (2)aa = 17.68 ms-2


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Diagram above shows a load of mass 500g ishung on a string C, which is tied to 2 other

strings A and B. Find the tension of string A.


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Tension of string C, TC = weight of the load = 5NAll forces in the system are in equilibrium, hence

Vertical component of tension A (TA) = TC

TAcos60o

= TCTA = TC/cos60o

TA = 5/cos60o = 10N

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Chapter 2.9 Equilibrium Presentation