# Q12. Static Equilibrium.

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Q12. Static Equilibrium

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>

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.

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>

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

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>

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

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>

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 :

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>

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° 

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>

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

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>

Movie