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Q12. Static Equilibrium

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**A picture can be hung on a wall in three different ways, as shown**

A picture can be hung on a wall in three different ways, as shown. The tension in the string is: least in I greatest in I greatest in II least in III greatest in III <PowerClick><Answer>5</Answer><Option>5</Option></PowerClick>

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**II is min. III is max. least in I greatest in I greatest in II**

least in III greatest in III II is min. III is max.

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A uniform plank XY is supported by two equal 120-N forces at X and Y, as shown. The support at X is then moved to Z (half-way to the plank center). The supporting forces at Y and Z are then:: FY = 240 N, FZ = 120 N FY = 200 N, FZ = 40 N FY = 40 N, FZ = 200 N FY = 80 N, FZ = 160 N FY = 160 N, FZ = 80 N <PowerClick><Answer>4</Answer><Option>5</Option></PowerClick>

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**Using Y as pivot : ( points upward ) FY = 240 N, FZ = 120 N**

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**A uniform rod AB is 1. 2 m long and weighs 16 N**

A uniform rod AB is 1.2 m long and weighs 16 N. It is suspended by strings AC and BD as shown. A block P weighing 96 N is attached at E, 0.30 m from A. The magnitude of the tension force in the string BD is: 8.0 N 24 N 32 N 48 N 80 N <PowerClick><Answer>3</Answer><Option>5</Option></PowerClick>

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1.2m T 0.3m 96N Using A as pivot : 8.0 N 24 N 32 N 48 N 80 N

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A 5.0 m weightless strut, hinged to a wall, is used to support a 800-N block as shown. The horizontal and vertical components of the force of the hinge on the strut are: FH = 800 N, FY = 800 N FH = 600 N, FY = 800 N FH = 800 N, FY = 600 N FH = 1200 N, FY = 800 N FH = 0 , FY = 800 N <PowerClick><Answer>2</Answer><Option>5</Option></PowerClick>

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**FY F FH Using hinge as pivot : FH = 800 N, FY = 800 N**

FH = 0 , FY = 800 N FY F FH Using hinge as pivot :

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**depends on the mass of the ladder depends on the length of the ladder**

A window washer attempts to lean a ladder against a frictionless wall. He finds that the ladder slips on the ground when it is placed at an angle of less than 75° to the ground but remains in place when the angle is greater than 75°. The coefficient of static friction between the ladder and the ground: is about 0.13 is about 0.27 is about 1.3 depends on the mass of the ladder depends on the length of the ladder <PowerClick><Answer>1</Answer><Option>5</Option></PowerClick>

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**L Total force = 0 Choose pivot point to be contact point on ground:**

is about 0.13 is about 0.27 is about 1.3 depends on the mass of the ladder depends on the length of the ladder L Total force = 0 Choose pivot point to be contact point on ground: N = N when 75°

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**The uniform rod shown below is held in place by the rope and wall**

The uniform rod shown below is held in place by the rope and wall. Suppose you know the weight of the rod and all dimensions. Then you can solve a single equation for the force exerted by the rope, provided you write expressions for the torques about the point: : 1 2 3 4 1, 2, or 3 <PowerClick><Answer>2</Answer><Option>5</Option></PowerClick>

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**Unknown forces are rope tension T & F2 . **

Choosing pivot point at 2 gives equation for one unknown T. 1 2 3 4 1, 2, or 3

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**When the truck moves to the right, the force(s) on**

Bridge and Truck: Movie When the truck moves to the right, the force(s) on the left scale increases. the right scale increases. both scales remain the same. <PowerClick><Answer>2</Answer><Option>3</Option></PowerClick>

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Movie

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Short Version : 12. Static Equilibrium. 12.1. Conditions for Equilibrium (Mechanical) equilibrium = zero net external force & torque. Static equilibrium.

Short Version : 12. Static Equilibrium. 12.1. Conditions for Equilibrium (Mechanical) equilibrium = zero net external force & torque. Static equilibrium.

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