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Force Vector Diagram #2

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Force Vector Diagram #2

A sign suspended by two ropes

When a body is suspended by two ropes as in the picture

above it has three forces acting upon it.

In the picture above F1 represents the left tension,

F2 represents the right tension, and F3 represents the

force of gravity.

FTL

FTR

Fg

The first thing that must be done when solving this sort of problem is to draw, as always, the force-vector diagram:

FTL stands for “force of tension on the left”, so what does FTR stand for?

FTL

FTR

FTL sin

FTR sin

FTL cos

FTR cos

Fg

Next, you must draw in the “components” of the vectors that are not completely aligned in the x and y directions:

(note that components are dotted and have arrowheads!)

As always, cosine is for the adjacent side, and sine is for the opposite side.

The hanging sign is in “equilibrium”, meaning that all of the forces balance each other out, creating an acceleration of zero in both dimensions!

FTL

FTR

FTL sin

FTR sin

FTL cos

FTR cos

Fg

So first add the forces in the x direction:

Fx = FTRcos – FTLcos = 0

Then add the forces in the y direction:

Fy = FTRsin + FTLsin – Fg = 0

Now all you’d have to do is solve the system of equations for FTR and FTL!

35°

25°

Ex 1: A 1000 kg picture is hanging from two ropes that form a 35 and a 25 degree angle with the horizontal. What are the tensions in both of the ropes?

FTR

FTL

FTR sin 25

FTL sin 35

25°

35°

FTR cos 25

FTL cos 35

Fg= 1000 x 9.8 = 9800N

Now try an example problem:

First, draw the force-vector-diagram:

FTR

FTL

FTR sin 25

FTL sin 35

25°

35°

FTR cos 25

FTL cos 35

constants on other side

of equation:

0

9800

FTR FTL

x eq: cos25 -cos35

y eq: sin25 sin35

-1

]

[

]

[

]

[

=

9269.58

10255.84

Fg= 1000 x 9.8 = 9800N

Sum the forces in the x direction:

Fx = FTRcos 25 – FTLcos 35 = 0

Then sum the forces in the y direction:

Fy = FTRsin 25 + FTLsin 35 – 9800 = 0

Use a matrix to solve:

Since FTR was the first variable in the matrix, then FTR 9300 N and FTL 10300 N (since FTL was the second variable in the matrix.).

FTL

FTRsin 40

FTR

FTRcos 40

FTLsin 60

40°

60°

FTLcos 60

Fg

=125N

Try another example ….

Note – on tonight’s HW, some angles are given from the vertical and some from the horizontal. If given from the vertical, you can do it the way it is done below, or just subtract the angle from 90 degrees to figure out the angle from the horizontal.

Ex 2: A no smoking sign is hanging by two ropes. The left rope forms a 60° angle with the horizontal and the right rope forms a 40° angle with the vertical. The sign weighs 125N. What are the tensions in the two ropes?

First make a diagram:

FTL

FTR

FTLsin

FTRsin

45°

60°

Fx =FTR sin40 FTL cos60 = 0

FTLcos

FTRcos

Fy =FTR cos40 + FTL sin60 125 = 0

Fg

=125N

1

]

[

]

[

]

[

=

66.5 N

85.5 N

sin40 -cos60

cos40 sin60

0

125

= FTR

= FTL

Then sum the forces:

And then solve:

Again – on tonight’s HW, for the angles given from the vertical, you can do it the way it is done above, or just subtract the angle from 90 degrees to figure out the angle from the horizontal.

Good Job

The End

Get the new HW ditto from the teacher, and do problems 1-15, drawing the FVD every time, just like last night.