MOMENT OF A COUPLE - Kwantlen Polytechnic University
MOMENT OF A COUPLE (continued) The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F d Moments due to couples can be added using the same rules as adding any vectors Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a free vector It
5 Moment of a Couple - Oakland University
The moment of the couple can be calculated using the cross product operator on the Matrix toolbar » M = cross(r_A,F_A) + cross(r_B,F_B); Newton centimeters » M = M/100 Newton meters M = 0 0 -120 0000 The result is a couple moment of 120 N⋅m directed in the –z direction (into the page) Solution 2
MOMENT OF A COUPLE - Philadelphia University
MOMENT OF A COUPLE (continued) Moments due to couples can be added together using the same rules as adding any vectors The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F *d Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a
MOMENT OF A COUPLE - DEU
MOMENT OF A COUPLE The moment of a couple is defined as MO = F d (using a scalar analysis) Important Note: Moment of a couple depends solely on the magnitude and the perpendicular distance between the forces, it is “free vector” A couple is defined as two parallel forces with the same magnitude but opposite in direction separated by a
5 MOMENTS, COUPLES, FORCES SYSTEMS & FORCE RESOLUTION
couple d F F F F Concept of a Couple When you grasp the opposite side of the steering wheel and turn it, you are applying a couple to the wheel A couple is defined as two forces (coplanar) having the same magnitude, parallel lines of action, but opposite sense Couples have pure rotational effectson the
ENGR-1100 Introduction to Engineering Analysis
MOMENT OF A COUPLE (continued) Moments due to couples can be added together using the same rules as adding any vectors The net external effect of a couple is that the net force equals zero and the magnitude of the net moment equals F *d Since the moment of a couple depends only on the distance between the forces, the moment of a couple is a
y SOLUTION - Florida International University
resultant couple moment Compute the result by resolving each force into x and y components and (a) finding the moment of each couple (Eq 4–13) and (b) summing the moments of all the force components about point A d = 4f t, 2f t B A y 1f t 3f t 50 lb 80 lb 50 lb 30˜ 30˜ 5 4 3 80 lb 3f t d x 5 4 3
More Examples on Moments
is the perpendicular distance between the lines of action of the couple forces, fig 3—30c However, the computation for d is more involved Realize that the couple moment is a free vector and can act at any point on the gear and produce the same turning effect about point O
13–1
the couple moment at the fixed support A Solution Equation of Motion Referring to the FBD of the crate shown in Fig a, +cΣF y = ma y; T - 50(9 81) = 50(6) T = 790 5 N Equations of Equilibrium Since the pulley is smooth, the tension is constant throughout entire cable Referring to the FBD of the pulley shown in Fig b, S+ ΣF x =0; 7 9
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245
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exist. No portion of this material may be reproduced, in any form or by �any means, without permission in writing from the publisher.
13-1.The 6-lb particle is subjected to the action of
its weight and forces and where tis in seconds.Determine the distance the ball is from the origin2safter being released from rest.F
3 =5-2ti6lb,5t 2 i-4tj-1k6lb,F 2 =F 1 =52i+6j-2tk6lb,SOLUTIONEquating components:
Since ,integrating from ,,yields
Since ,integrating from ,yields
When then,,
Thus,Ans.s=
(14.31) 2 +(35.78) 2 +(-89.44) 2 =97.4fts z =-89.44fts y =35.78fts x =14.31ftt=2s 6 32.2s x =t4 12 -t 3 3+t 2 6 32.2
s y =-2t 3 3+3t 2 6 32.2
s z =-t 3 3-7t 2
2t=0s=0ds=vdt
6 32.2v x =t 3 3-t 2 +2t 6 32.2
v y =-2t 2 +6t 6 32.2
v z =-t 2 -7tt=0n=0dv=adt 6 32.2
a x =t 2 -2t+2 6 32.2
a y =-4t+6 6 32.2
a z =-2t-7©F=ma;(2i+6j-2tk)+(t 2 i-4tj-1k)-2ti-6k= 6 32.2
(a x i+a y j+az k z y x F 1 F 3 F 2Ans: s=97.4 ft 247
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exist. No portion of this material may be reproduced, in any form or by �any means, without permission in writing from the publisher.
13-3. If the coefficient of kinetic friction between the 50-kg crate and the ground is ,determine thedistance the crate travels and its velocity when The crate starts from rest,and .P=200 Nt=3 s.m k =0.3SOLUTION
Free-Body Diagram:The kinetic friction is directed to the left to oppose the motion of the crate which is to the right,Fig.a.Equations of Motion:Here,.Thus,
Kinematics:Since the acceleration aof the crate is constant, A n s. and A n s.s=0+0+1 2 (1.121)A3 2B=5.04 ms=s
0 +v 0 t+1 2 a c t 2 A:+B v=0+1.121(3)=3.36 m>sv=v 0 +a c tA:+B a=1.121 m>s 2200 cos 30°-0.3(390.5)=50a©F
x =ma x:+N=390.5 NN-50(9.81)+200 sin 30°=0+c©F y =0a y =0F f =m k N 30P Ans: v =3.36 m>s s=5.04 m 248
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*13-4. If the 50-kg crate starts from rest and achieves a velocity of when it travels a distance of 5 m to the right, determine the magnitude of force Pacting on thecrate. The coefficient of kinetic friction between the crate and the g round is .m k =0.3v=4 m>sSOLUTION
Kinematics:The acceleration aof the crate will be determined first since its motion i s known. Free-Body Diagram:Here,the kinetic friction is required to be directed to the left to oppose the motion of the crate which is to the right,Fig.a.Equations of Motion:
U s in g the results of Nand a, A ns.P=224 NP cos 30°-0.3(490.5-0.5P)=50(1.60):+©F
x =ma xN=490.5-0.5PN+P
sin 30°-50(9.81)=50(0)+c©F y =ma y F f =m kN=0.3Na=1.60 m>s
2 :4 2 =0 2 +2a(5-0)v 2 =v 0 2 +2a c s-s 0 30P Ans:
P=224 N
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exist. No portion of this material may be reproduced, in any form or by �any means, without permission in writing from the publisher.
13-6.The 10-lb block has a speed of 4 ft
>s when the force of F =(8t 2 ) lb is applied. Determine the velocity of the block when t=2 s. The coefficient of kinetic friction at the surface is m k =0.2.SOLUTION
Equations of Motion.
Here the friction is F
f =m kN=0.2N. Referring to the FBD
of the block shown in Fig. a +cF y =ma y ; N-10= 10 32.2(0) N=10 lb S+F x =ma x ; 8t 2 -0.2(10)= 10 32.2
a a =3.22(8t 2 -2) ft>s 2
Kinematics.
The velocity of the block as a function of
t can be determined by integrating d=a dt using the initial condition =4 ft>s at t=0. L4 ft>s
d=L t 03.22 (8t
2 -2)dt -4=3.22 a8 3 t 3 -2tb =5 -+6ft>s When t=2 s, =59.81 ft>s =59.8 ft>s Ans. v 4 ft/s F (8 t 2 ) lb Ans: =59.8 ft>s 259© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights r�eserved. This material is protected under all copyright laws as they cur�rently
exist. No portion of this material may be reproduced, in any form or by �any means, without permission in writing from the publisher.