Pre-Lab:The following figure is anairplane's view of a force table;a 120 gram masshangs off of pulley at 400. A second mass of 500 grams hangs off of anotherpulley at 205° according to the picture in Figure 6.1. The weights pull on thering in the center of the table via the pulleys.Picturein33A. Use vector components to find the combined total force (magnitude and direction)exerted on the ring by the two cables with weights hanging off of them.B. At what angle should a 3rd pulley be placed, and how much mass hanging off of it wouldbe needed to balance the system in part A so that the central ring does not move? fieureAirplane ViewForce TablePulley 1205°400Side ViewPulley 1Pulley 2CentralRingSide ViewPulley 2120grams500Airplane View of Force TablegramsFORCH FABLE/ouldFigure 6.1: Force Table

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Asked Oct 3, 2019
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I need help with part A and B. 

Pre-Lab:
The following figure is an
airplane's view of a force table;
a 120 gram mass
hangs off of pulley at 400. A second mass of 500 grams hangs off of another
pulley at 205° according to the picture in Figure 6.1. The weights pull on the
ring in the center of the table via the pulleys.
Picturein
33
A. Use vector components to find the combined total force (magnitude and direction)
exerted on the ring by the two cables with weights hanging off of them.
B. At what angle should a 3rd pulley be placed, and how much mass hanging off of it would
be needed to balance the system in part A so that the central ring does not move?
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Pre-Lab: The following figure is an airplane's view of a force table; a 120 gram mass hangs off of pulley at 400. A second mass of 500 grams hangs off of another pulley at 205° according to the picture in Figure 6.1. The weights pull on the ring in the center of the table via the pulleys. Picturein 33 A. Use vector components to find the combined total force (magnitude and direction) exerted on the ring by the two cables with weights hanging off of them. B. At what angle should a 3rd pulley be placed, and how much mass hanging off of it would be needed to balance the system in part A so that the central ring does not move?

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fieure
Airplane View
Force Table
Pulley 1
205°
400
Side View
Pulley 1
Pulley 2
Central
Ring
Side View
Pulley 2
120
grams
500
Airplane View of Force Table
grams
FORCH FABLE
/ould
Figure 6.1: Force Table
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fieure Airplane View Force Table Pulley 1 205° 400 Side View Pulley 1 Pulley 2 Central Ring Side View Pulley 2 120 grams 500 Airplane View of Force Table grams FORCH FABLE /ould Figure 6.1: Force Table

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Expert Answer

Step 1

A.

Consider the vector diagram given below.

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2050 400 F

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Step 2

The forces are as follows.

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F = mg (0.120kg(9.8m/s =1.176N Fm2g (0.500kg)(9.8m/s = 4.9N

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Step 3

Forces in vector ...

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(1.176Ncos 40+ sin 40°j] =(0,901N)+(0.756N) F =(4.9Ncos 205+ sin 205°] =(4.441N)+(2071N)

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Vectors and Scalars

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