Engineering Mechanics - Force Vectors - Discussion
Discussion Forum : Force Vectors - General Questions (Q.No. 4)
4.
Determine the magnitude and direction of the resultant force.
R = 80.3 lb, 
= 106.2° CCW
= 106.2° CCWR = 80.3 lb, = 73.8° CCW
= 73.8° CCWR = 72.1 lb, = 63.6° CCW
= 63.6° CCWR = 72.1 lb, = 116.4° CCW
= 116.4° CCWDiscussion:
13 comments Page 1 of 2.
Bright said:
8 years ago
Why is 60cos45 negative?
Toms said:
8 years ago
How would you solve the x and y components if you don't know the angles?
Thato said:
8 years ago
@All.
Try parallelogram method to get the answer.
Try parallelogram method to get the answer.
Sarvjeet singh said:
9 years ago
R = 80.25 s/ direction = 73.72.
(2)
Krishna said:
9 years ago
@Mani Obul.
Yes, why you subtracted 73.7 from 180?
Yes, why you subtracted 73.7 from 180?
Bob said:
10 years ago
Why do you have to minus 73.7 from 180?
Ng kl said:
1 decade ago
I calculated the answer is -7.78.
(60 lb) (cos 45) is negative.
(60 lb) (cos 45) is negative.
D.mani obul reddy said:
1 decade ago
Horizontal components = 60cos45 - 40sin30 = 22.426.
Vertical components = 60sin45+40cos30 = 77.067.
R = SQUARE ROOT(22.426)^2+(77.067)^2 = 80.26.
ANGLE = TAN inverse of(77.067/22.426) = 73.77.
180 - 73.77 = 106.2 ccw.
Vertical components = 60sin45+40cos30 = 77.067.
R = SQUARE ROOT(22.426)^2+(77.067)^2 = 80.26.
ANGLE = TAN inverse of(77.067/22.426) = 73.77.
180 - 73.77 = 106.2 ccw.
(1)
Cj manglalan said:
1 decade ago
CCW I THINK IT MEANS IT ROTATES Counter ClockkWise
FOR F:
F1x= (60 lb) (cos 45)= 42.426
f1y= 42.426 (sin 45=cos45) or (60 lb)(sin 45)
FOR P:
p1x= (40lb)(sin 30) = 20
p1y= (40lb)(cos 30) =34.641
sumamtion of forces along x =42.426 + 20 = -22.426
summation of forces along y =42.426 + 34.641 = 77.067
for R:
R= square root of (((-22.426)square)+ ((77.067)square))
for angle:
=arc tangent (y/x)
=arc tangent (77.067/-22.426) = -73.8
since the angle is negative you will subtract it to 180
the angle is
180-73.8 = 106.2
FOR F:
F1x= (60 lb) (cos 45)= 42.426
f1y= 42.426 (sin 45=cos45) or (60 lb)(sin 45)
FOR P:
p1x= (40lb)(sin 30) = 20
p1y= (40lb)(cos 30) =34.641
sumamtion of forces along x =42.426 + 20 = -22.426
summation of forces along y =42.426 + 34.641 = 77.067
for R:
R= square root of (((-22.426)square)+ ((77.067)square))
for angle:
=arc tangent (y/x)
=arc tangent (77.067/-22.426) = -73.8
since the angle is negative you will subtract it to 180
the angle is
180-73.8 = 106.2
Snehasish said:
1 decade ago
CCW means angle makes from west corner (-ve X-axis) i.e. 180 - 73.77 = 106.228 ~ 106.2.
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