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9. Resolve the force into rectangular components with the given co-ordinate axes as shown in Fig. E. (Ans. (a) F_x = -F sin ?, F_y = F cos ? (b) F_x = F sin ?, F_y = F cos ? (c) F_x = F sin ?; F_y = -F cos ? (d) F_x = F cos (? - ?); F_y = F sin (? - ?) Fig. E 10. In level flight, the chord AB of an airplane wing makes an angle ? = 5° with the horizontal (Fig. F). The resultant wind pressure on the wing for such conditions is defined by its lift and drag components L = 6675 N and D = 890 N, which are vertical and horizontal, respectively, as shown. Resolve this force into rectangular components X and Y, coinciding with the chord AB and its normal, respectively. (Ans. X=304.8 N; Y = 6727 N) Fig. F 11. A small block of weight Q = 44.5 N is placed on an inclined plane which makes an angle ? = 30° with the horizontal. Resolve the gravity force Q into two rectangular components Q_t and Q_n acting parallel and normal, respectively, to the inclined plane. (Ans. Q_t = 22.25 N; Q_n = 38.54 N)

          9. Resolve the force into rectangular components with the given co-ordinate axes as shown in Fig. E.
(Ans. (a) F_x = -F sin ?, F_y = F cos ?
(b) F_x = F sin ?, F_y = F cos ?
(c) F_x = F sin ?; F_y = -F cos ?
(d) F_x = F cos (? - ?); F_y = F sin (? - ?)
Fig. E
10. In level flight, the chord AB of an airplane wing makes an angle ? = 5° with the horizontal (Fig. F). The resultant wind pressure on the wing for such conditions is defined by its lift and drag components L = 6675 N and D = 890 N, which are vertical and horizontal, respectively, as shown. Resolve this force into rectangular components X and Y, coinciding with the chord AB and its normal, respectively.
(Ans. X=304.8 N; Y = 6727 N)
Fig. F
11. A small block of weight Q = 44.5 N is placed on an inclined plane which makes an angle ? = 30° with the horizontal. Resolve the gravity force Q into two rectangular components Q_t and Q_n acting parallel and normal, respectively, to the inclined plane.
(Ans. Q_t = 22.25 N; Q_n = 38.54 N)

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9. Resolve the force into rectangular components with the given co-ordinate axes as shown in Fig. E.
(Ans. (a) Fx = -F sin ?, Fy = F cos ?
(b) Fx = F sin ?, Fy = F cos ?
(c) Fx = F sin ?; Fy = -F cos ?
(d) Fx = F cos (? - ?); Fy = F sin (? - ?)
Fig. E
10. In level flight, the chord AB of an airplane wing makes an angle ? = 5° with the horizontal (Fig. F). The resultant wind pressure on the wing for such conditions is defined by its lift and drag components L = 6675 N and D = 890 N, which are vertical and horizontal, respectively, as shown. Resolve this force into rectangular components X and Y, coinciding with the chord AB and its normal, respectively.
(Ans. X=304.8 N; Y = 6727 N)
Fig. F
11. A small block of weight Q = 44.5 N is placed on an inclined plane which makes an angle ? = 30° with the horizontal. Resolve the gravity force Q into two rectangular components Qt and Qn acting parallel and normal, respectively, to the inclined plane.
(Ans. Qt = 22.25 N; Qn = 38.54 N)

Added by Tara S.

University Physics with Modern Physics

Hugh D. Young 14th Edition

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Solved by Expert Stephen Zaffke

12/31/2022

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We need to resolve this force into rectangular components X and Y, coinciding with the chord AB and its normal, respectively. From the given information, we can see that the lift component L is vertical and the drag component D is horizontal. Show more…

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Resolve the force into rectangular components with the given coordinate axes as shown in Fig: (Ans: (a) F_x = Fsinα, F_y = Fcosα & (b) F_x = Fsinα, F_y = Fcosα & (c) F_x = Fsinα, F_y = -Fcosα & (d) F_x = Fcos(α-β), F_y = Fsin(α-β)) Fig: E 10. In level flight, the chord AB of an airplane wing makes an angle α = 5° with the horizontal (Fig F). The resultant wind pressure on the wing for such conditions is defined by its lift and drag components L = 6675 N and D = 890 N, which are vertical and horizontal, respectively, as shown. Resolve this force into rectangular components X and Y, coinciding with the chord AB and its normal, respectively: (Ans. X = 304.8 N; Y = 6727 N) Fig: F 11. A small block of weight Q = 44.5 N is placed on an inclined plane which makes an angle α = 30° with the horizontal. Resolve the gravity force Q into two rectangular components Q_t and Q_n acting parallel and normal, respectively, to the inclined plane. (Ans: Q_t = 22.25 N; Q_n = 38.54 N)

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Transcript

00:01 All right, so there's actually three problems listed here.

00:04 So what i'll do is i'll answer the first one, and hopefully you can resubmit the other two and get answers for those as well.

00:13 So for this first problem, we have four different coordinate systems, and we need to divide this force into its x and y components.

00:21 So this first one here, letter a, we have the force at an angle alpha with respect to the y axis.

00:28 And so we need to divide this into its coordinate parts.

00:33 And the opposite side, so we're going to draw the legs of a right triangle around this force.

00:39 And the angle here is going to give us the x component, which is the opposite side to it, and the y component, which is the adjacent side.

00:48 So the x component here will be in the negative x direction.

00:52 So we'll write f subx is equal to negative.

00:56 And then it's the magnitude of the force, which is our hypotenuse here, times the sign of this angle.

01:03 So f sign of alpha is that.

01:07 And again, this is in the negative direction, so that's why i've drawn the negative sign there.

01:12 And the y component here will be with the cosine.

01:16 So again, f cosine of theta, or cosine of alpha, i should say.

01:24 And in this case, we're in the positive y direction.

01:27 Stays positive.

01:30 For letter b, we can do much the same where now we have the positive x direction is pointed down on our axis here.

01:39 But again, we're just dividing this into its x and my components.

01:44 And so right here we have the positive x direction is down.

01:47 Positive y direction is to the left.

01:50 And so this alpha angle here, if we look at this opposite side, that's going to give us the positive of x direction.

01:58 So f sub x is going to be f sine of alpha and f some y is going to be the adjacent side.

02:09 So that gives us f cosine of alpha like so.

02:16 Okay...

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