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

Force-Vector-Diagram #6 The very last one! Yeah!. F N. Table Mass. F T. F T. F f = F N. Hanging Mass. F gT. F gH. Force-Vector-Diagram #6

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The very last one!

Yeah!

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.

FN

Table Mass

FT

FT

Ff = 0.700FN

Hanging Mass

FgT = 1421 N

FgH = 3969 N

Example Problem

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!