Mechanical Engineering Principles

FRICTION 175

plane is so much smaller than to move the body up
the plane.

15.7 Motion up a plane due to a

horizontal forceP

This motion, together with the primary forces, is
shown in Figure 15.8.
In this, the components of mg are as shown in
Figure 15.4, and the components of the horizon-
tal forceP are shown by the vector diagram of
Figure 15.9.

q

q

Motion

N

P

F

mg

Figure 15.8

q

q

q

P sin

q

P cos

q

P

Plane

Figure 15.9

Resolving perpendicular to the plane gives:

Forces ‘up’ =forces ‘down’

i.e. N=mgcosθ+Psinθ( 15. 11 )

Resolving parallel to the plane gives:

Pcosθ=F+mgsinθ( 15. 12 )

and F=μN ( 15. 13 )

From equations (15.11) to (15.13), problems arising
in this category can be solved.

Problem 9. If the mass of Problem 7 were

subjected to a horizontal forceP, as shown

in Figure 15.8, determine the value ofP that

will just cause motion up the plane.

Substituting equation (15.13) into equation (12)

gives:

Pcosθ=μN+mgsinθ

or μN=Pcosθ−mgsinθ

i.e. N=

Pcosθ

μ

−

mgsinθ

μ

( 15. 14 )

Equating equation (15.11) and equation (15.14)

gives:

mgcosθ+Psinθ=

Pcosθ

μ

−

mgsinθ

μ

i.e. 25× 9 .81 cos 15°+Psin 15°

=

Pcos 15°

0. 3

−

25 × 9 .81 sin 15°

0. 3

245. 3 × 0. 966 +P× 0. 259

=

P× 0. 966

0. 3

−

245. 3 × 0. 259

0. 3

i.e. 237 + 0. 259 P= 3. 22 P− 211. 8

237 + 211. 8 = 3. 22 P− 0. 259 P

from which, 448. 8 = 2. 961 P

and forceP=

448. 8

2. 961

= 151 .6N

Problem 10. If the mass of Problem 9 were

subjected to a horizontal forceP, acting

down the plane, as shown in Figure 15.10,

determine the value ofPwhich will just

cause motion down the plane.

q

q

N

P F

mg

Motion

Figure 15.10