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Force-Vector-Diagram #6 The very last one! Yeah!PowerPoint Presentation

Force-Vector-Diagram #6 The very last one! Yeah!

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FN

Table Mass

FT

FT

Ff = FN

Hanging Mass

FgT

FgH

Force-Vector-Diagram #6

A mass is resting on a table top. There is a string attached, which runs over a pulley on the edge of the table, and is attached to another “hanging” mass.

The equations below are derived from the force-vector diagram on the previous slide. The “hanging” or “table” indicate on which object the forces were acting. To get the equations, simply add the forces for each object in the y and then in the x. This gives you three equations (because there really is no x for the hanging mass).

Table: FyT = FN – FgT = 0

Table: FxT = -Ff + FT = mTa

Hanging: FyH = FgH – FT = mHa

Solving the Problem will require a matrix (I wouldn’t copy this down, as we’ll see it used in the actual example that’s next!)

FT a #

-1

FyH

FxT

-1-mH

1-mT

-FgH

Ff

The first matrix is setup using the coefficients of FT and a.

In the second matrix, FgH represents the force of gravity or weight of the “hanging” mass and Ff represents the force of friction created by the “table” mass sliding across the table.

Take the inverse A-1B to solve.

FT a #

-1

FyH

FxT

-1-405

1-145

-3969

994.7

Plugging in the Numbers

Table: FyT = FN – 1421 = 0; so FN = 1421N

Table: FxT = – (.7)(1421) + FT = (1421/9.8)a

Hanging: FyH = 3969 – FT = (3969/9.8)a

Remember – there are LOTS of correct matrices!

Answers to the Problem:

FT = 1780 N

a = 5 .41 m/s2

The diagram described in force-vector-diagram type 6 can also play a part in the real world. Say someone falls into a canyon and is badly injured. Help must be lowered down to that person to rescue them. The amount of friction at the top of the canyon and the masses of the two objects attached to the rope play roles in determining how many people it will take to hold the rope at the top of the canyon and how heavy the weight being lowered down can be.

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