How do you find the resultant of these two forces?
Q. Two forces, 481 N at 21 deg and 371 N at 30 deg are applied to a car in an effort to acclerate it. A) find the resultant of these two forces B) Find the direction of the resultant force in units of deg C) If the car has no mass of 3600 kg, what acceleration does it have? in units of m/s^2
Asked by ruby - Tue Sep 29 00:22:08 2009 - - 2 Answers - 0 Comments
A. I will assume your angles are measured counterclocwise upward from the positive x axis. a) the first force has components x1: 481 cos 21= 449N y1: 481 sin 21 = 172.4N the second force has components x2=371cos30=321.3N y2=674sin30=185.5N the components of the resultant are: x=x1+X2=449+321.3=770.3N y=y1+y2=172.4+185.5=357.9 N magnitude of resultant=sqrt[770.3^2+35 7.9^2] = 849.4N b) direction of resultant is given by tan (theta) = y component/x component = 357.9/770.3 theta = arctan(0.46) so theta = 24.9 deg c) the total force is 849.4N if this acts on a car of 3600kg, the acceleration is determined from F=ma or a=F/m = 849.4N/3600kg =0.24m/s/s
Answered by kuiperbelt2003 - Tue Sep 29 00:38:16 2009
Q. Two forces, 481 N at 21 deg and 371 N at 30 deg are applied to a car in an effort to acclerate it. A) find the resultant of these two forces B) Find the direction of the resultant force in units of deg C) If the car has no mass of 3600 kg, what acceleration does it have? in units of m/s^2
Asked by ruby - Tue Sep 29 00:22:08 2009 - - 2 Answers - 0 Comments
A. I will assume your angles are measured counterclocwise upward from the positive x axis. a) the first force has components x1: 481 cos 21= 449N y1: 481 sin 21 = 172.4N the second force has components x2=371cos30=321.3N y2=674sin30=185.5N the components of the resultant are: x=x1+X2=449+321.3=770.3N y=y1+y2=172.4+185.5=357.9 N magnitude of resultant=sqrt[770.3^2+35 7.9^2] = 849.4N b) direction of resultant is given by tan (theta) = y component/x component = 357.9/770.3 theta = arctan(0.46) so theta = 24.9 deg c) the total force is 849.4N if this acts on a car of 3600kg, the acceleration is determined from F=ma or a=F/m = 849.4N/3600kg =0.24m/s/s
Answered by kuiperbelt2003 - Tue Sep 29 00:38:16 2009
Find the angle between the forces given the magnitude of their resultant?
Q. Force 1 = 45 pounds Force 2 = 60 pounds Resultant force = 90 pounds (hint: write force 1 as a vector in the direction of the positive x-axis and force 2 as a vector at an angle theta with the positive x-axis) <--(what the book suggested...)
Asked by Andrew T - Wed Oct 31 20:54:32 2007 - - 2 Answers - 0 Comments
A. F1= 45 i F2= 60 cos t*i+60 sin t *j F1+F2 = (45+60cos t) i+60 sin t j (45+60 cos t)^2 +(60 sin t) ^2 = 8100 2025 +3600 cos^2t +5400 cos t +3600 sin^t =8100 5625 +5400cost = 8100 cos t=0.4583 so t=62.72
Answered by santmann2002 - Wed Oct 31 21:43:53 2007
Q. Force 1 = 45 pounds Force 2 = 60 pounds Resultant force = 90 pounds (hint: write force 1 as a vector in the direction of the positive x-axis and force 2 as a vector at an angle theta with the positive x-axis) <--(what the book suggested...)
Asked by Andrew T - Wed Oct 31 20:54:32 2007 - - 2 Answers - 0 Comments
A. F1= 45 i F2= 60 cos t*i+60 sin t *j F1+F2 = (45+60cos t) i+60 sin t j (45+60 cos t)^2 +(60 sin t) ^2 = 8100 2025 +3600 cos^2t +5400 cos t +3600 sin^t =8100 5625 +5400cost = 8100 cos t=0.4583 so t=62.72
Answered by santmann2002 - Wed Oct 31 21:43:53 2007
What is the resultant force for these two forces?
Q. One is 400 Newtons at 30 degrees north of east and the other is 295 Newtons at 40 degrees south of east.
Asked by sadfasdgsadfsadf - Sun Dec 2 17:23:22 2007 - - 1 Answers - 0 Comments
A. x=400cos30+295cos40 572.3932 y=400sin30-295sin40=10.37 76 (resultant)^2=x^2+y^2 (resultant)^2=(572.3932)^ 2+(10.3776)^2 resultant=572.4873N
Answered by tiger_vision2036 - Sun Dec 2 17:36:37 2007
Q. One is 400 Newtons at 30 degrees north of east and the other is 295 Newtons at 40 degrees south of east.
Asked by sadfasdgsadfsadf - Sun Dec 2 17:23:22 2007 - - 1 Answers - 0 Comments
A. x=400cos30+295cos40 572.3932 y=400sin30-295sin40=10.37 76 (resultant)^2=x^2+y^2 (resultant)^2=(572.3932)^ 2+(10.3776)^2 resultant=572.4873N
Answered by tiger_vision2036 - Sun Dec 2 17:36:37 2007
forces of 250 and 450 newtons, forming an angle of 85 degrees find the magnitude and the resultant force.?
Q. I need this explained as well as answered. Thanks.
Asked by George - Sun Mar 30 22:48:55 2008 - - 1 Answers - 0 Comments
A. This will be a matter of vector addition. To simplify the problem a little, I'll place one of the vectors, with tail at the origin, flush against the x-axis. That way it has no vertical component we have to calculate. So let's put the 450N vector on the x-axis. Its tail will touch the origin. Its head will be at the point (450,0). Then stick the tail of the 250N vector at (450,0), and draw the rest of it pointing toward the upper right, so that it forms an 85 angle with the x-axis. The resultant force will be represented by the vector that connects the tail of the 450N vector at (0,0) to the head of the 250N vector... i.e. we're just doing vector addition, i.e. vector A + vector B = vector C. Now we just have to find the x-componen [cont.]
Answered by Charles D - Mon Mar 31 12:46:50 2008
Q. I need this explained as well as answered. Thanks.
Asked by George - Sun Mar 30 22:48:55 2008 - - 1 Answers - 0 Comments
A. This will be a matter of vector addition. To simplify the problem a little, I'll place one of the vectors, with tail at the origin, flush against the x-axis. That way it has no vertical component we have to calculate. So let's put the 450N vector on the x-axis. Its tail will touch the origin. Its head will be at the point (450,0). Then stick the tail of the 250N vector at (450,0), and draw the rest of it pointing toward the upper right, so that it forms an 85 angle with the x-axis. The resultant force will be represented by the vector that connects the tail of the 450N vector at (0,0) to the head of the 250N vector... i.e. we're just doing vector addition, i.e. vector A + vector B = vector C. Now we just have to find the x-componen [cont.]
Answered by Charles D - Mon Mar 31 12:46:50 2008
find by graphical means the resultant force of two forces 15N n 20N 60 degress to each other?
Q. i got 30N in the end am i right?
Asked by Noddy - Sat Jan 30 08:53:38 2010 - - 3 Answers - 0 Comments
Q. i got 30N in the end am i right?
Asked by Noddy - Sat Jan 30 08:53:38 2010 - - 3 Answers - 0 Comments
when two forces act on an object in the same direction, the resultant force is equal to....?
Q. a. the sum of the forces b. the difference of the forces c. the product of the forces d. the ratio of the forces
Asked by ashley. - Thu Mar 11 16:07:04 2010 - - 6 Answers - 0 Comments
A. a) the sum of the forces
Answered by David - Thu Mar 11 16:12:03 2010
Q. a. the sum of the forces b. the difference of the forces c. the product of the forces d. the ratio of the forces
Asked by ashley. - Thu Mar 11 16:07:04 2010 - - 6 Answers - 0 Comments
A. a) the sum of the forces
Answered by David - Thu Mar 11 16:12:03 2010
Two forces act on an object at an angle of 50 degrees. One force is 150 N. The resultant force is 200 N.?
Q. Find the second force and the angle it makes with the resultant. Can use Cartesian vector methods ONLY. Any help is appreciated. Full methodology preferred. Thanks in advance.
Asked by FollowHunterS - Sat Nov 21 16:37:49 2009 - - 1 Answers - 0 Comments
Q. Find the second force and the angle it makes with the resultant. Can use Cartesian vector methods ONLY. Any help is appreciated. Full methodology preferred. Thanks in advance.
Asked by FollowHunterS - Sat Nov 21 16:37:49 2009 - - 1 Answers - 0 Comments
For the 3 forces in the diagram , the magnitude and the direction of the resultant force are?
Q. 1) 110N , 0 2) 70 N , 90 3) 50 N , 53.1 4) 50 N , 36.9 what is after the comma is the angle the digram is in that link , please i need the answer with some dealits
Asked by Tarkan - Sun Jun 22 01:47:39 2008 - - 4 Answers - 0 Comments
A. Since 20N and 50N are collinear and opposite, we can get their resultant: -20 N + 50N = + 30N (due to the right) Now, the resultant of + 30 N due to the right and 40 N due upward is... Use the Pythagorean Theorem: R = sqrt[30^2 + 40^2] R = 50N = angle = arctan(40/30) = 53.1 deg Therefore, R = 50N, 53.1 deg north of east answer answer: 3 Hope this helps. teddy boy
Answered by teddy boy - Sun Jun 22 05:22:19 2008
Q. 1) 110N , 0 2) 70 N , 90 3) 50 N , 53.1 4) 50 N , 36.9 what is after the comma is the angle the digram is in that link , please i need the answer with some dealits
Asked by Tarkan - Sun Jun 22 01:47:39 2008 - - 4 Answers - 0 Comments
A. Since 20N and 50N are collinear and opposite, we can get their resultant: -20 N + 50N = + 30N (due to the right) Now, the resultant of + 30 N due to the right and 40 N due upward is... Use the Pythagorean Theorem: R = sqrt[30^2 + 40^2] R = 50N = angle = arctan(40/30) = 53.1 deg Therefore, R = 50N, 53.1 deg north of east answer answer: 3 Hope this helps. teddy boy
Answered by teddy boy - Sun Jun 22 05:22:19 2008
2 forces F & G act parallel to the vectors 3i+j & i+2j respectively.If the resultant is a force of 10 newtons?
Q. Con.> in the direction of the vector 3i-4j,find the magnitudes of F & G in surd form.Details: i,j,F &G are in bold format..itz mechanic question on resultant forces..Thnkz
Asked by Spirit - Mon Jul 28 10:04:37 2008 - - 1 Answers - 0 Comments
A. let F = a(3i + j) and G = b(i+2j) F + G = a(3i+j) + b(i+2j) = (3a + b)i + (a +2b)j let resultant be R then R = 10(3i -4j)/(sqrt(3^2 + 4^2)) = 6i - 8j now F + G = R so, (3a + b)i + (a +2b)j = 6i -8j so, 3a+b = 6 and a+2b = -8 solving these we get a = 4,b = -6 F = 4(3i+j) = 12i+4j G = -6(i+2j) = -6i - 12j magnitude of F = sqrt(12^2 + 4^2) = sqrt(160) = 4*sqrt(10) magnitude of G = sqrt(6^2 + 12^2) = sqrt(180) = 6*sqrt(5)
Answered by Anubarak - Fri Aug 1 07:17:55 2008
Q. Con.> in the direction of the vector 3i-4j,find the magnitudes of F & G in surd form.Details: i,j,F &G are in bold format..itz mechanic question on resultant forces..Thnkz
Asked by Spirit - Mon Jul 28 10:04:37 2008 - - 1 Answers - 0 Comments
A. let F = a(3i + j) and G = b(i+2j) F + G = a(3i+j) + b(i+2j) = (3a + b)i + (a +2b)j let resultant be R then R = 10(3i -4j)/(sqrt(3^2 + 4^2)) = 6i - 8j now F + G = R so, (3a + b)i + (a +2b)j = 6i -8j so, 3a+b = 6 and a+2b = -8 solving these we get a = 4,b = -6 F = 4(3i+j) = 12i+4j G = -6(i+2j) = -6i - 12j magnitude of F = sqrt(12^2 + 4^2) = sqrt(160) = 4*sqrt(10) magnitude of G = sqrt(6^2 + 12^2) = sqrt(180) = 6*sqrt(5)
Answered by Anubarak - Fri Aug 1 07:17:55 2008
Forces??????????????????? ???????????
Q. "Two forces of 130 and 150 pounds yield a resultant force of 170 pounds. Find, to the nearest tenth of a degree, the angle between the original two forces." Ehh...please help me. A step by step explanation would be extremely appreciative as I'd like to understand how this problem is done rather than just getting an answer... :i
Asked by September - Tue Dec 22 21:10:14 2009 - - 1 Answers - 0 Comments
A. You can use the law of cosines to determine the angle between the vectors c2 = a^2 + b^2 - 2*a*b*cos(theta) So theta = arccos(( a^2 + b^2 -c^2)/(2*a*b)) = arccos(130^2 + 150^2 - 170^2)/(2*130*150)) = 74.4o
Answered by DH - Tue Dec 22 21:22:12 2009
Q. "Two forces of 130 and 150 pounds yield a resultant force of 170 pounds. Find, to the nearest tenth of a degree, the angle between the original two forces." Ehh...please help me. A step by step explanation would be extremely appreciative as I'd like to understand how this problem is done rather than just getting an answer... :i
Asked by September - Tue Dec 22 21:10:14 2009 - - 1 Answers - 0 Comments
A. You can use the law of cosines to determine the angle between the vectors c2 = a^2 + b^2 - 2*a*b*cos(theta) So theta = arccos(( a^2 + b^2 -c^2)/(2*a*b)) = arccos(130^2 + 150^2 - 170^2)/(2*130*150)) = 74.4o
Answered by DH - Tue Dec 22 21:22:12 2009
Mechanics Resultant forces help?
Q. Hi I have a diagram of four concurrent forces acting in different directions with angles in between and need to find the following resultant horizontal force magnitude and direction of of overall force am i right in saying that for the horizontal force i take the 2 forces on the right hand side and add them then cos the 2 angles? Any help is appreciated!
Asked by Tommy - Sat Jan 16 12:02:32 2010 - - 1 Answers - 0 Comments
A. 0. Impose an x-y coordinate system: positive x is horizontal to the right, positive y is up, etc. 1. For each force, a. Compute its direction as angle measured counterclockwise relative to the positive x axis. b. Decompose the force into x and y components (f * cos(a), f * sin(a)) 2. Add the x components together and add the y components together. This gives the resultant vector as an x-y pair. 3. The absolute value x value is the resultant horizontal force. Because force is a vector, magnitude and direction, a negative value does make sense. It would mean the force is directed to the left. 4. The magnitude of the resultant force is sqrt(x^2 + y^2). To get the direction, I recommend making a quick sketch so you can see which quadrant… [cont.]
Answered by Mike T - Sat Jan 16 12:57:00 2010
Q. Hi I have a diagram of four concurrent forces acting in different directions with angles in between and need to find the following resultant horizontal force magnitude and direction of of overall force am i right in saying that for the horizontal force i take the 2 forces on the right hand side and add them then cos the 2 angles? Any help is appreciated!
Asked by Tommy - Sat Jan 16 12:02:32 2010 - - 1 Answers - 0 Comments
A. 0. Impose an x-y coordinate system: positive x is horizontal to the right, positive y is up, etc. 1. For each force, a. Compute its direction as angle measured counterclockwise relative to the positive x axis. b. Decompose the force into x and y components (f * cos(a), f * sin(a)) 2. Add the x components together and add the y components together. This gives the resultant vector as an x-y pair. 3. The absolute value x value is the resultant horizontal force. Because force is a vector, magnitude and direction, a negative value does make sense. It would mean the force is directed to the left. 4. The magnitude of the resultant force is sqrt(x^2 + y^2). To get the direction, I recommend making a quick sketch so you can see which quadrant… [cont.]
Answered by Mike T - Sat Jan 16 12:57:00 2010
How do you find the direction and magnitude of the resultant force, that is, how do you find F1 F2?
Q. Two forces, F1, of magnitude 60 newtons (N) and F2 of magnitude 70 newtons, act on an object at angles of 40 degrees and 130 degrees (respectively) with the positive x-axis. Find the direction and magnitude of the resultant force, that is, find F1+F2. Round the direction and magnitude to two decimal places. How do I do this? This is the only problem on my precalc study guide that I do not get. Please help!
Asked by adamcohen13 - Mon Feb 1 12:37:34 2010 - - 1 Answers - 0 Comments
A. first you need to find the x and y components F1x = cos 40 x 60N = 45.96 F1y = sin 40 x 60N = 38.57 F2x = cos 130 x 70N= -44.99 F2y = sin 130 x 70N = 53.62 Now you add the x and y components square each of them and find the square root that will give you 92.20N that is the magnitude Now for the direction you get the addition of the x and y components and find the tan^-1 y/x and that will give you the direction in degrees = 89.40 more or less Hope it helps
Answered by jesus - Mon Feb 1 13:22:14 2010
Q. Two forces, F1, of magnitude 60 newtons (N) and F2 of magnitude 70 newtons, act on an object at angles of 40 degrees and 130 degrees (respectively) with the positive x-axis. Find the direction and magnitude of the resultant force, that is, find F1+F2. Round the direction and magnitude to two decimal places. How do I do this? This is the only problem on my precalc study guide that I do not get. Please help!
Asked by adamcohen13 - Mon Feb 1 12:37:34 2010 - - 1 Answers - 0 Comments
A. first you need to find the x and y components F1x = cos 40 x 60N = 45.96 F1y = sin 40 x 60N = 38.57 F2x = cos 130 x 70N= -44.99 F2y = sin 130 x 70N = 53.62 Now you add the x and y components square each of them and find the square root that will give you 92.20N that is the magnitude Now for the direction you get the addition of the x and y components and find the tan^-1 y/x and that will give you the direction in degrees = 89.40 more or less Hope it helps
Answered by jesus - Mon Feb 1 13:22:14 2010
How to find resultant and equilibrant force?
Q. Three forces are exerted on a point. First one is 6.0 N north; second is 35 N east; third is 40 N at 30 degrees east of south. What is the resultant force and equilibrant force? the answer is 62N, 27 degrees south of east for resultant force; 62N, 27 degrees North of West, but I have no clue how to get those answers.
Asked by TS - Thu Oct 8 02:05:20 2009 - - 1 Answers - 0 Comments
A. You need to break each of the force vectors into their respective x and y components then add like components. F1 = 6.0 N north or in the positive y direction F2 = 35N east or in the positive x direction F3 = 40N at 30 east of south meaning at an angle of 300 degrees. Okay lets find forces in the same direction X Direction: F1x = 0 F2x = 35N F3x = 40cos300 Frx = 0 + 35 + (40cos30) = 55N Y Direction F1y = 6.0N F2y = 0N F3y = 40sin300 Fry = 6.0 + 0 + 40sin300 = -28.64N Fr = sqrt (Frx^2 + Fry^2) = sqrt (55^2 + (-20.64^2)) = 62.01N Angle (k) of the vector tan(k) = Fry / Frx = -28.64 / 55 k = tan^-1( -28.64 / 55) = -27.50 meaning that its a 27.50 degree angle subtending from the x axis or 27.50 degrees south of east.
Answered by Michael - Thu Oct 8 02:27:21 2009
Q. Three forces are exerted on a point. First one is 6.0 N north; second is 35 N east; third is 40 N at 30 degrees east of south. What is the resultant force and equilibrant force? the answer is 62N, 27 degrees south of east for resultant force; 62N, 27 degrees North of West, but I have no clue how to get those answers.
Asked by TS - Thu Oct 8 02:05:20 2009 - - 1 Answers - 0 Comments
A. You need to break each of the force vectors into their respective x and y components then add like components. F1 = 6.0 N north or in the positive y direction F2 = 35N east or in the positive x direction F3 = 40N at 30 east of south meaning at an angle of 300 degrees. Okay lets find forces in the same direction X Direction: F1x = 0 F2x = 35N F3x = 40cos300 Frx = 0 + 35 + (40cos30) = 55N Y Direction F1y = 6.0N F2y = 0N F3y = 40sin300 Fry = 6.0 + 0 + 40sin300 = -28.64N Fr = sqrt (Frx^2 + Fry^2) = sqrt (55^2 + (-20.64^2)) = 62.01N Angle (k) of the vector tan(k) = Fry / Frx = -28.64 / 55 k = tan^-1( -28.64 / 55) = -27.50 meaning that its a 27.50 degree angle subtending from the x axis or 27.50 degrees south of east.
Answered by Michael - Thu Oct 8 02:27:21 2009
How do you find the magnitude and direction of a resultant force?
Q. Hi, I am unsure how to go about finding the magnitude, direction and line of action of the resultant of the two forces acting on the rigid frame shown in the link below: Could you please show your working. Cheers
Asked by magic_ian - Wed Aug 5 01:43:55 2009 - - 1 Answers - 0 Comments
A. Find the x and y components of each force, add them, giving the components of the resultant: Components of the 800 N force: x component = 0 y component = 800 N Components of the 600 N force x component = 600*cos y component = -600*sin From the little triangle given for the angle, tan = 0..5 = 26.5 x component =600*cos26.5 = 536.7 N y component = -600*/ 5 = -268.3 N Sum the components sum x = 536.7 sum y = 800 - 268.3 = 531.7 Magnitude of resultant = [x + y ] = [536.7 + 531.7 ] = 755 N The angle is arctan y/x = arctan 531.7/536.7 = 44.7 I'm not sure what "line of action" means, but Wikipedia says that is the same as the angle. However, it may mean the location of a single force with the same effect as the two… [cont.]
Answered by gp4rts - Wed Aug 5 04:55:54 2009
Q. Hi, I am unsure how to go about finding the magnitude, direction and line of action of the resultant of the two forces acting on the rigid frame shown in the link below: Could you please show your working. Cheers
Asked by magic_ian - Wed Aug 5 01:43:55 2009 - - 1 Answers - 0 Comments
A. Find the x and y components of each force, add them, giving the components of the resultant: Components of the 800 N force: x component = 0 y component = 800 N Components of the 600 N force x component = 600*cos y component = -600*sin From the little triangle given for the angle, tan = 0..5 = 26.5 x component =600*cos26.5 = 536.7 N y component = -600*/ 5 = -268.3 N Sum the components sum x = 536.7 sum y = 800 - 268.3 = 531.7 Magnitude of resultant = [x + y ] = [536.7 + 531.7 ] = 755 N The angle is arctan y/x = arctan 531.7/536.7 = 44.7 I'm not sure what "line of action" means, but Wikipedia says that is the same as the angle. However, it may mean the location of a single force with the same effect as the two… [cont.]
Answered by gp4rts - Wed Aug 5 04:55:54 2009
What is the x component and y component of the resultant force?
Q. Two forces F_1 and F_2, act at a point. F_1 has a magnitude of 9.00 and is directed at an angle of 63.0 degrees above the negative x axis in the second quadrant. F_2 has a magnitude of 5.20 and is directed at an angle of 53.7 degrees below the negative x axis in the third quadrant.
Asked by bg - Wed Sep 26 12:47:18 2007 - - 1 Answers - 2 Comments
A. vector f = mag(f) * (cos(theta),sin(theta)) when theta is CCW relative to +x axis. f1 = (-4.085914D+000 +8.019059D+000) f2 = (-3.078469D+000 -4.190827D+000) ft = f1+f2 = (-7.164383D+000 +3.828232D+000) Magnitude (ft) = 8.123038 Angle (CCW from +x) = arctan2(v3(2), v3(1)) = 151.8825 deg arctan2(Vy, Vx) accommodates the full 360-deg range of angles whereas the normal arctan(Vy/Vx) function can only output angles in the +x half of the plane, -90 to 90 deg. In this case it's helpful since the problem was set up for a -x resultant.
Answered by kirchwey - Wed Sep 26 13:15:51 2007
Q. Two forces F_1 and F_2, act at a point. F_1 has a magnitude of 9.00 and is directed at an angle of 63.0 degrees above the negative x axis in the second quadrant. F_2 has a magnitude of 5.20 and is directed at an angle of 53.7 degrees below the negative x axis in the third quadrant.
Asked by bg - Wed Sep 26 12:47:18 2007 - - 1 Answers - 2 Comments
A. vector f = mag(f) * (cos(theta),sin(theta)) when theta is CCW relative to +x axis. f1 = (-4.085914D+000 +8.019059D+000) f2 = (-3.078469D+000 -4.190827D+000) ft = f1+f2 = (-7.164383D+000 +3.828232D+000) Magnitude (ft) = 8.123038 Angle (CCW from +x) = arctan2(v3(2), v3(1)) = 151.8825 deg arctan2(Vy, Vx) accommodates the full 360-deg range of angles whereas the normal arctan(Vy/Vx) function can only output angles in the +x half of the plane, -90 to 90 deg. In this case it's helpful since the problem was set up for a -x resultant.
Answered by kirchwey - Wed Sep 26 13:15:51 2007
What is the x component fx of the resultant force?
Q. Two forces, and , act at a point, as shown in the picture. has a magnitude of 8.60 and is directed at an angle of = 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.40 and is directed at an angle of = 52.5 below the negative x axis in the third quadrant.
Asked by Naveta F - Sat Feb 9 21:27:18 2008 - - 2 Answers - 0 Comments
A. 8.6 * cos((90 + 56) degrees) + 5.4 * cos((180+52.5) degrees) = -10.4170348
Answered by antiver - Sat Feb 9 21:32:30 2008
Q. Two forces, and , act at a point, as shown in the picture. has a magnitude of 8.60 and is directed at an angle of = 56.0 above the negative x axis in the second quadrant. has a magnitude of 5.40 and is directed at an angle of = 52.5 below the negative x axis in the third quadrant.
Asked by Naveta F - Sat Feb 9 21:27:18 2008 - - 2 Answers - 0 Comments
A. 8.6 * cos((90 + 56) degrees) + 5.4 * cos((180+52.5) degrees) = -10.4170348
Answered by antiver - Sat Feb 9 21:32:30 2008
if the two forces act on an object, is the resultant force equal to the sum of the two forces?
Q. second question... how is the resultant of two forces affected by the angle between the force? 3rd question... if a force is exerted to pull an object, is it the whole force that makes the object move??? Please Specify your source. Also make your answer understandable by a 14 year old kid...
Asked by georgekb81 - Tue Jul 3 08:11:59 2007 - - 2 Answers - 0 Comments
A. 1. Yes, if the two forces are co-linear (in other words, acting in the same direction). If they aren't co-linear, see #2. 2. If the two forces are not co-linear, then you need to compare the component of the forces along the direction of interest, usually the x and y axes in Cartesian coordinates. You can find these components using trigonometry. For example, if you have a force F acting at 30 deg from the horizontal, the component of the force in the horizontal direction is F cos 30 and in the vertical direction F sin 30. 3. Yes, the net force, in other words.
Answered by TR - Tue Jul 3 08:17:04 2007
Q. second question... how is the resultant of two forces affected by the angle between the force? 3rd question... if a force is exerted to pull an object, is it the whole force that makes the object move??? Please Specify your source. Also make your answer understandable by a 14 year old kid...
Asked by georgekb81 - Tue Jul 3 08:11:59 2007 - - 2 Answers - 0 Comments
A. 1. Yes, if the two forces are co-linear (in other words, acting in the same direction). If they aren't co-linear, see #2. 2. If the two forces are not co-linear, then you need to compare the component of the forces along the direction of interest, usually the x and y axes in Cartesian coordinates. You can find these components using trigonometry. For example, if you have a force F acting at 30 deg from the horizontal, the component of the force in the horizontal direction is F cos 30 and in the vertical direction F sin 30. 3. Yes, the net force, in other words.
Answered by TR - Tue Jul 3 08:17:04 2007
Resultant Forces question?
Q. I know its an extremely easy problem. Just want to clarify. There are two forces acting from a point A. The two forces extend outwards in order to form a 90 degree angle. One force extend at 0 degrees north and the other force at 90 degrees to the east. Show the direction and the size of the force required to balance point A.
Asked by Hellow there xD - Mon Jun 2 04:02:09 2008 - - 2 Answers - 0 Comments
Q. I know its an extremely easy problem. Just want to clarify. There are two forces acting from a point A. The two forces extend outwards in order to form a 90 degree angle. One force extend at 0 degrees north and the other force at 90 degrees to the east. Show the direction and the size of the force required to balance point A.
Asked by Hellow there xD - Mon Jun 2 04:02:09 2008 - - 2 Answers - 0 Comments
one force of 20pounds and one force of 15 pounds act on a body at the same point so that the resultant forces.
Q. one force of 20pounds and one force of 15 pounds act on a body at the same point so that the resultant forces.
Asked by creamiexhunie - Tue May 6 18:34:32 2008 - - 1 Answers - 0 Comments
A. ?
Answered by ~Calli~ - Tue May 6 18:42:59 2008
Q. one force of 20pounds and one force of 15 pounds act on a body at the same point so that the resultant forces.
Asked by creamiexhunie - Tue May 6 18:34:32 2008 - - 1 Answers - 0 Comments
A. ?
Answered by ~Calli~ - Tue May 6 18:42:59 2008
Knowing that the resultant of the two forces shown must be horizontal, determine the angle (greg letter)?
Q. knowing that the resultant of the two forces shown must be horizontal, determine the angle (greek letter) and the magnitude of the resultant force. can somebody explain me how to solve this. thanks
Asked by azulita - Wed Jan 23 11:52:11 2008 - - 1 Answers - 0 Comments
A. Your link is broken but if theta = 0 then the sum of the y compontents must = 0 becasue the arctan of 0 = 0 Decompose each vector into is x and y components Sum the y components, then sum the x components Pay close attention to the direction of the compontents, are they negative or positive.( going up down left or right) Magnitude of the vector = sqrt(x^+y^2) Theta=arctan(y/x)
Answered by raymond w - Wed Jan 23 12:02:18 2008
Q. knowing that the resultant of the two forces shown must be horizontal, determine the angle (greek letter) and the magnitude of the resultant force. can somebody explain me how to solve this. thanks
Asked by azulita - Wed Jan 23 11:52:11 2008 - - 1 Answers - 0 Comments
A. Your link is broken but if theta = 0 then the sum of the y compontents must = 0 becasue the arctan of 0 = 0 Decompose each vector into is x and y components Sum the y components, then sum the x components Pay close attention to the direction of the compontents, are they negative or positive.( going up down left or right) Magnitude of the vector = sqrt(x^+y^2) Theta=arctan(y/x)
Answered by raymond w - Wed Jan 23 12:02:18 2008
From Yahoo Answer Search: 'resultant of forces'
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