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Q12. Static Equilibrium. 1.A picture can be hung on a wall in three different ways, as shown. The tension in the string is: 1. least in I 2. greatest.

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Presentation on theme: "Q12. Static Equilibrium. 1.A picture can be hung on a wall in three different ways, as shown. The tension in the string is: 1. least in I 2. greatest."— Presentation transcript:

1 Q12. Static Equilibrium

2 1.A picture can be hung on a wall in three different ways, as shown. The tension in the string is: 1. least in I 2. greatest in I 3. greatest in II 4. least in III 5. greatest in III

3  II is min. III is max. 1. least in I 2. greatest in I 3. greatest in II 4. least in III 5. greatest in III

4 2.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:: 1. F Y = 240 N, F Z = 120 N 2. F Y = 200 N, F Z = 40 N 3. F Y = 40 N, F Z = 200 N 4. F Y = 80 N, F Z = 160 N 5. F Y = 160 N, F Z = 80 N

5 Using Y as pivot : ( points upward ) 1. F Y = 240 N, F Z = 120 N 2. F Y = 200 N, F Z = 40 N 3. F Y = 40 N, F Z = 200 N 4. F Y = 80 N, F Z = 160 N 5. F Y = 160 N, F Z = 80 N 

6 3.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: N N N N N

7 Using A as pivot : 1.2m 0.3m N N N N N T 96N

8 4.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: 1. F H = 800 N, F Y = 800 N 2. F H = 600 N, F Y = 800 N 3. F H = 800 N, F Y = 600 N 4. F H = 1200 N, F Y = 800 N 5. F H = 0, F Y = 800 N

9 Using hinge as pivot :  1. F H = 800 N, F Y = 800 N 2. F H = 600 N, F Y = 800 N 3. F H = 800 N, F Y = 600 N 4. F H = 1200 N, F Y = 800 N 5. F H = 0, F Y = 800 N FHFH FYFY F

10 5.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: 1. is about is about is about depends on the mass of the ladder 5. depends on the length of the ladder

11 Total force = 0  Choose pivot point to be contact point on ground: N =  N when   75°  1. is about is about is about depends on the mass of the ladder 5. depends on the length of the ladder L

12 6.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: : , 2, or 3

13 , 2, or 3 Unknown forces are rope tension T & F 2. Choosing pivot point at 2 gives equation for one unknown T. T

14 7.Bridge and Truck: 1. the left scale increases. 2. the right scale increases. 3. both scales remain the same. Movie When the truck moves to the right, the force(s) on

15 Movie


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