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Physics For Scientists & Engineers
Found in: Page 355

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Short Answer

Calculate the net torque (magnitude and direction) on the beam in Figure P11.5 about

(a) an axis through O perpendicular to the page and

(b) an axis through C perpendicular to the page.

a) Therefore, the net torque about the point O perpendicular is 30.0N.m

The direction of the net torque is in counter clockwise direction

b) Therefore, the net torque about perpendicular is 36.0N.m

The direction of the net torque is in counter clockwise direction

See the step by step solution

Step by Step Solution

Step 1: Formula to calculate the Torque

τ=rF

Here is the Torque, is the perpendicular force and is the applied force.

The force that acts on the beam is shown in figure

From this figure OOis the perpendicular distance for the forceF1 ,OD is the perpendicular distance for the forceF2,OO is the perpendicular distance for the forceF3 ,F1cosθ1 is the horizontal component of the forceF1 ,F2sinθ2 is the vertical component ofF2 ,F3sinθ3 is the vertical component of the forceF3

Step 2: To find the net torque about perpendicular point

The net torque about the point O perpendicular is

τO=OCF1cosθ1+OD(-F2sinθ2)+OO(-F3sinθ3)

τOis the net torque about the point O,OO is the perpendicular distance for the forceF1 ,OD is the perpendicular distance for the forceF2 ,OO is the perpendicular distance from the F3, F1cosθ1is the horizontal component for F1,F2sinθ2 is the horizontal component for F2,F3sinθ3 is the horizontal component forF3

Substitute 2.0m for OC,25N for F1,30o for θ1,4.0m for OD,10N for F2,20o for θ2,0m for OO,30N for F3,and 45o for θ,intheequation

τO=OCF1cosθ1+OD(-F2sinθ2)+OO(-F3sinθ3) andsolvefor τO

τO=(2.0m)(25N)cos30o-(4.0m)(10N)sin20o-(0m)(30N)sin45o

τO=30.0N×m

Therefore, the net torque about the point O perpendicular is30.0N×m

The direction of the net torque is in counter clockwise direction

The net torque about the point C perpendicular is

τC=(CC)F1sinθ1+(OC)(-F2sinθ2)+(-OC)(-F3sinθ3)

τCis the net torque about the point C,CC is the perpendicular distance for the forceF1

Substitute 0m for CC,25N for F1,30o for θ1,2.0m for OC,10N for F2,20o for θ2

30N for F,and 45o for θ,intheequation

τC=(CC)F1sinθ1+(OC)(-F2sinθ2)+(-OC)(-F3sinθ3) andsolvefor τC

τC=(0m)(25N)cos30o-(2.0m)(10N)sin20o+(2.0m)(30N)sin45o

τC=36.0N×m

Therefore, the net torque about perpendicular is36.0N×m

The direction of the net torque is in counter clockwise direction

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