ME 232 Classroom Problems

February 4, 1998

Box Resting on Inclined Plane

The figure shows a uniform rectangular box supported by a roller on its lower left corner on an inclined plane.  The box has width L = 6 ft and height H = 4 ft.  The box has weight W=1000 lb.  The inclined plane makes an angle q with the horizontal such that tan(q) = 3/4 as shown.  The box is released from rest in the position shown.  Determine the initial acceleration of the roller down the incline.  Determine the initial normal force exerted by the incline on the roller.

0) Think:  As we are asked to analyze the system at the instant shown, we will need to sum forces and moments and set equal to the appropriate kinematic terms.  As the system is released from rest, all of the initial velocity terms will be zero.

1) Mechanical System = Box plus roller at the instant shown.

2) Free Body Diagram.  Note the use of two coordinate systems.  The XY system is horizontal and vertical while the xy system is parallel to and perpendicular to the incline.  The x and X axes make an angle of q with each other.  Both axes are useful in this problem.  Also note that as the roller provides no resistance to moving down the incline and no resistance to rotation of the box, the only force acting at the roller is normal (perpendicular) to the inclined surface.  Recalling that the slope of the perpendicular is the negative reciprocal of the original slope, this normal force acts along a line with slope 4/3.

3) Equations:
a)KineticsS F = N j - W J
= N j - W { cos(q) j - sin(q) i }
= W sin(q) i + { N - W cos(q) } j

As the box does not have a fixed point, we must sum moments about the mass center.
S MG = { -L/2 I - H/2 J ) X N j
= { -L/2 I - H/2 J ) X N { cos(q) J + sin(q) I }
= { -N L/2 cos(q) + N H/2 sin(q) } K
= { -N L/2 cos(q) + N H/2 sin(q) } k
b)Kinematics: The acceleration of the mass center can be related to the acceleration of the roller using the two point formula.  In the following a subscript R indicates the roller.
m aG = (W/g) [ aR + w X ( w X RG/R) + a X RG/R
Noting that the angular velocity is zero at the instant of release and that the roller moves along the incline:
m aG  = { W/g } [ aR i + 0 + a k X ( L/2 I + H/2 J) ]
= { W/g } [ aR i + a L/2 J - a H/2 I ]
= { W/g } [ aR i + a L/2 ( cos(q) j - sin(q) i ) - a H/2 (cos(q) i + sin(q) j ) ]
= { W/g } [ (aR - a L/2 sin(q) - a H/2 cos(q) ) i + ( a L/2 cos (q) - a H/2 sin(q) ) j ]
The angular mass of the box is that of a uniform rectangle:
IG a = 1/12 { W/g } [ L2 + H2 } a k

4) Solve:
W sin(q) = { W/g }  (aR - a L/2 sin(q) - a H/2 cos(q) )     x force equation.
N - W cos(q) = { W/g }( a L/2 cos (q) - a H/2 sin(q) )      y force equation.
-N L/2 cos(q) + N H/2 sin(q) =  1/12 { W/g } [ L2 + H2 ] a       z moment equation.

The above represents three equations in the three unknowns: N, aR and a.  The bottom equation can be solved for the angular acceleration in terms of the normal force.  This expression for the normal force can then be plugged into the second equation which can then be solved for the normal force.  The expression developed for the angular acceleration in terms of the normal force can then be used to evaluate the angular acceleration.  With the normal force and the angular acceleration known, the top equation can then be solved for the acceleration of the roller.  Solving the bottom equation for the angular acceleration in terms of the normal force:

a = 12 { g/W } 1/[L2+H2] {-L/2 cos(q) + H/2 sin(q } N

Plugging back into the second equation:

N - W cos(q) = { W/g }( L/2 cos (q) -  H/2 sin(q) )  12 { g/W } 1/[L2+H2] {-L/2 cos(q) + H/2 sin(q } N

Solving for N:

N = W cos(q) / ( 1 + 12/[L2+H2( L/2 cos (q) -  H/2 sin(q) )2 }

Using the known values for L, H, W, and the angle:

N = 600 lb

Now we can determine the value of the angular acceleration: