Problem 3/18 Page 132, Engineering Mechanics - Dynamics, Meriam and Kraige, 4th Edition

Given: A sphere of mass ms suspended as shown in a frame of mass mf is sliding down an incline making an angle q with the horizontal.  The incline is rough with coefficient of kinetic friction mk.  The supporting wires each make an angle of qf with the edge of the frame parallel to the incline.

Find: The tension in the wires supporting the sphere.

Parameter Values:
ms = 10 kg
mf = 15 kg
q = 20 degrees
mk = 0.15
qf = 45 degrees

1. Mechanical System:  Frame + Sphere + Support Wires at instant shown.

2. Free Body Diagram: Not included.  Would show frame, sphere, support wires, total weight force acting vertically downward, normal force acting perpendicular to the incline, and a friction force acting up the incline (resisting the motion).  The friction force is equal to the normal force multiplied by the kinetic coefficient of friction.  The coordinate axes are shown adjacent to the free body diagram.  The t direction points down the incline.  The n direction is perpendicular to the incline.

3. Equations:
S F = { (mf + ms) g sin q - mk N } t + { N - (mf + ms) g cos(q) } n
m a = (mf + ms) { at t + 0 n }

4. Solve:
Noting that there is no acceleration in the n direction:
N - (mf + ms) g cos(q) = 0
Solving for the normal force:
N = (mf + ms) g cos(q)
Considering the t direction:
(mf + ms) g sin q - mk N = (mf + ms) at
Using the determined value of the normal force:
(mf + ms) sin q - mk (mf + ms) g cos(q) = (mf + ms) at
Dividing through by the total mass and solving for at:
at = g {sin q - mk cos(q) }

1. Mechanical System:  Sphere  at instant shown.

2. Free Body Diagram: Not included.  Would show sphere, sphere weight force acting vertically downward, and tensions in the two wires.  The coordinate axes are shown adjacent to the free body diagram.  The t direction points down the incline.  The n direction is perpendicular to the incline.

3. Equations:
S F = { ms g sin q - TA cos qf + TB cos qf } t + { TA sin qf + TB sin qf -  ms g cos(q) } n
m a =  ms { at t + 0 n } = ms g { sin q - mk cos(q) }t

4. Solve:
Considering the t and n directions yields two equations in the two unknown tensions, TA and TB:
ms g sin q - TA cos qf + TB cos qf = ms g { sin q - mk cos(q) }
TA sin qf + TB sin qf -  ms g cos(q) = 0

Putting the equations in standard form:

-cos qf  TA + cos qf   TB = -ms g  mk cos(q)
  sin qf   TA+  sin qf    TB =   ms g cos(q)

Multiplying the top equation by sin qf and the bottom equation by cos qf and adding:

2 cos qf sin qf   TB = ms g cos q { cos qf - mk sin qf }

Solving for TB
TB = ms g cos q { cos qf- mk sin qf } / { 2 cos qf sin qf } = 55.4 N

Multiplying the top equation by -sin qf and the bottom equation by cos qf and adding:

2 cos qf sin qf   TA = ms g cos q { cos qf + mk sin qf }

Solving for TA
TA = ms g cos q { cos qf + mk sin qf } / { 2 cos qf sin qf } = 75.0 N