Engineering Mechanics

Chapter 9 : Applications of Friction „„„„„ 167

We know that as there are two threads in a cm, (i.e. n = 2) therefore pitch of the screw,
p = 1/2 = 0.5 cm
and internal diameter of the screw,
= 5 – (2 × 0.5) = 4 cm
∴ Mean diameter of the screw,
54
4.5 cm
2

d

+

==

Let α = Helix angle.

We know that

0.5

tan 0.0353

4.5

p

d

α= = =

ππ×

(i) Force required at the end of 70 cm long lever to lift the load

Let P 1 = Force required at the end of the lever to lift the load.

We know that the force required to be applied at the mean radius to lift the load,

tan tan

tan ( )

1– tan .tan

PW W

α+ φ

=α+φ=×

α φ

0.0353 0.1

4000 543.1 N

1 – 0.0353 0.1

+

=× =

×

Now the force required at the end of the lever may be found out from the relation,

(^1)
4.5
70 543.1 1222
22
d
PP×=×= × =
∴ 1
1222
17.5 N
7.0
P== Ans.
(ii) Force required at the end of 70 cm long lever to lower the load
Let P 2 = Force required at the end of the lever to lower the load.
We know that the force required at the mean radius to lower the load,
tan – tan
tan ( – ) 4000
1tantan
PW
φ α
=φα=×

• φ α
  0.1 – 0.0353
  4000 257.9 N
  1 0.1 0.0353
  =× =
  +×
  Now the force required at the end of the lever may be found out from the relation:
  2
  4.5
  70 257.9 580.3
  22
  d
  PP×=×= × =
  ∴ 2
  580.3
  8.3 N
  70
  P== Ans.
  9.7. EFFICIENCY OF A SCREW JACK
  We have seen in Art. 9.6 that the effort (P) required at the mean radius of a jack to lift the
  load (W),
  P = W tan (α + φ) ...(i)
  where α = Helix angle, and
  μ = Coefficient of friction between the screw and the nut,
  = tan φ ...(where φ = Angle of friction)