Mechanical Engineering Principles

174 MECHANICAL ENGINEERING PRINCIPLES

For limiting friction,

F=μN ( 15. 5 )

From equations (15.3) to (15.5), solutions of
problems in this category that involve limiting
friction can be solved.

Problem 7. Determine the value of the

forceP, which will just move the body of

mass of 25 kg up the plane shown in

Figure 15.6. It may be assumed that the

coefficient of limiting friction,μ= 0 .3and

g= 9 .81 m/s^2.

Motion

P

F q

N

15 ° mg

Figure 15.6

From equation (15.4),

N= 25 × 9. 81 ×cos 15°

= 245. 3 × 0. 966 = 236 .9N

From equation (15.5),

F= 0. 3 × 236. 9 = 71 .1N

From equation (15.3),

P= 25 × 9. 81 ×sin 15°+ 71. 1

= 63. 48 + 71. 1

i.e. force,P= 134 .6N ( 15. 6 )

15.6 Motion down a plane with the

pulling forcePparallel to the

plane

In this case, the frictional forceFacts up the plane,
opposite to the direction of motion of the plane, as
shown in Figure 15.7.
The components of the weight mg are shown in
Figure 15.4, where it can be seen that the normal
reaction, N = mgcosθ, and the component of
weight parallel to and down the plane=mgsinθ.

q

q

Motion

Plane

F

P

N

mg

Figure 15.7

Resolving forces perpendicular to the plane gives:

N=mgcosθ( 15. 7 )

Resolving forces parallel to the plane gives:

P+mgsinθ=F( 15. 8 )

When the friction is limiting,

F=μN ( 15. 9 )

From equations (15.7) to (15.9), problems arising in

this category can be solved.

Problem 8. If the mass of Problem 7 were

subjected to the forceP, which acts parallel

to and down the plane, as shown in

Figure 15.7, determine the value ofPto just

move the body.

From equation (15.7),

N= 25 × 9 .81 cos 15°= 236 .9N

From equation (15.9),

F= 0. 3 × 236. 9 = 71 .1N

From equation (15.8),

P+ 25 × 9 .81 sin 15°= 71. 1

i.e. P+ 63. 5 = 71. 1

from which,

force,P= 71. 1 − 63. 5 = 7 .6N ( 15. 10 )

From equations (15.6) and (15.10), it can be seen

that the force required to move a body down the