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
The very last one!
Ff = FN
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 #
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.
Ff = 0.700FN
FgT = 1421 N
FgH = 3969 N
FT a #
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.