Mechanical Engineering Principles

176 MECHANICAL ENGINEERING PRINCIPLES

q

q

P

P sin q

P cos q

Figure 15.11

The components formg are shown by the phasor
diagram of Figure 15.4, and the components forP
are shown by the vector diagram of Figure 15.11.

Resolving forces down the plane gives:

Pcosθ+mgsinθ=F( 15. 15 )

Resolving forces perpendicular to the plane gives:

Forces up=forces down

N+Psinθ=mgcosθ( 15. 16 )

and F=μN ( 15. 17 )

Substituting equation (15.17) into equation (15.15)
gives:

Pcosθ+mgsinθ=μN

from which, N=

Pcosθ

μ

+

mgsinθ

μ

( 15. 18 )

From equation (15.16),

N=mgcosθ−Psinθ( 15. 19 )

Equating equations (15.18) and (15.19) gives

Pcosθ

μ +

mgsinθ

μ =mgcosθ−Psinθ

i.e. Pcos 15

°

0. 3 +

25 × 9 .81 sin 15°

0. 3

= 25 × 9 .81 cos 15°−Psin 15°

3. 22 P+ 211. 6 = 236. 9 − 0. 259 P

P( 3. 22 + 0. 259 )= 236. 9 − 211. 6

3. 479 P= 25. 3

from which, forceP=

25. 3

3. 479

= 7 .27 N

Problem 11. If in Problem 9, the contact

surfaces were greased, so that the value ofμ

decreased andP=50 N, determine the

value ofμwhich will just cause motion

down the plane.

The primary forces for this problem are shown in

Figure 15.12, where it can be seen thatFis opposite

to the direction of motion.

F

N

mg

P= 50 N

Motion

q

q

Figure 15.12

Resolving forces perpendicular to the plane gives:

Forces ‘up’=forces ‘down’

N=mgcosθ+Psinθ( 15. 20 )

Resolving forces parallel to the plane gives:

mgsinθ=F+Pcosθ( 15. 21 )

and F=μN ( 15. 22 )

Substituting equation (15.22) into equation (15.21)

gives:

mgsinθ=μN+Pcosθ( 15. 23 )

Substituting equation (15.20) into equation (15.23)

gives:

mgsinθ=μ(mgcosθ+Psinθ)+Pcosθ

i.e. 25× 9 .81 sin 15°=μ( 25 × 9 .81 cos 15°

+50 sin 15°)+50 cos 15°

Hence 63. 48 =μ( 236. 89 + 12. 94 )+ 48. 3

63. 48 − 48. 3 =μ× 249. 83

from which, μ=

15. 18

249. 83

= 0. 061