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Mechanical principles assignment

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This is a while assignment to achieve a Merit standard in the Mechanical principles Unit in Level 3 Engineering year . All to do with turning moments

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  • December 28, 2014
  • 9
  • 2014/2015
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h4rn
B
8KN
Task 2a (P1*)
30°


5KN
A
A 4KN
C 1m



45°


0.5m

3KN
D


0.5m 1m


Determine the magnitude, direction and sense of the resultant and equilibrant forces acting on the
above component and the perpendicular distance of their line of action from point A.

Solution:


A. Fv= F×cos×ϑ A. Fh= F×sin×ϑ B. Fh= F×cos×ϑ B. Fv= F×sin×ϑ
= 4×cos×45 = 4×sin×45 = 8×cos×30 = 8×sin×30
= 2.8KN = 2.8KN = 6.9KN = 4KN


Fh=2.8KN 4KN Fv=4KN
8KN
A B




Fv=2.8KN Fh=6.9KN




Forces Fh Fv Turning moments Turning moments
3KN +3 0 0 3 3KN -4.2 0
4KN +2.8 +2.8 4.2 0 4KN 6.9 -2
5KN 0 5 0 7.5 5KN 0 0
8KN -6.9 +4 3.4 2 8KN 0 -1.5

∑= -1..8

,To find theta:

ϑ=cos-1( )


=cos-1( )

=84.6°
FR=11.74KN
Use of Pythagoras theorem to find resultant force (R). Fv=11.8

FR=

=139.24-1.21

=√

=11.74
ϑ
Fh=-1.1


To find the turning moments i multiplied the length from the pivot and the force

Forces Fh Fv
3KN -1.5 0
4KN 0 -4.2
5KN 0 0
8KN -2 -6.9
-3.5 -11.1

∑m-14.6Nm


x=

x=

=-1.2




The arrow going downwards is the
equilibrant force, which has the
same values as the Resultant force

, Task 2b (P1*)

A crate of mass of 50-100 Kg is held in equilibrium by a force T acting parallel to the plane
as indicated. Determine using the resolution method the magnitude of the force T and the
normal reaction R ignoring the effects of friction.


R

P


80KG




30°




Gravity=9.81N

w = mg

w = 80x9.81

w = 784.8N

(To find "R" we use cos and to find "P" we use sin)

R= w×cos×ϑ P= w×sin×ϑ

R=784.8×cos×30 P=784.8×sin×30

R=679.65N P=392.4N

The force "P" is 392.4N and the reaction force "R" is 679.65N.

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