2.3 Rectangular Components in Three-Dimensional Force Systems

Transcription

1 2.3 Rectangular Components in Three-Dimensional Force Sstems

2 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 1, page 1 of 2 1. Epress the force F in terms of,, and components. F = 200 lb 45 2 Calculate the component. F = 200 lb = = View of plane formed b the ais and F. F = 200 lb d 5 The component points in the negative direction, so 4 d = (200 lb) cos 60 = 100 lb F = 100 lb Ans. 120

3 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 1, page 2 of 2 6 View of plane formed b the ais and F 7 component F = 200 lb F = (200 lb) cos 45 = lb Ans. 45 F 8 View of plane formed b the ais and F. F = 200 lb 9 component F = (200 lb) cos 60 = 100 lb Ans. F F = Fi + F j + Fk = { 100i j + 100k} lb Ans.

4 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 2, page 1 of 2 2. Epress F in terms of,, and components. F = 50 N A B 1 View of plane formed b A, F, and the ais. 2 F = (50 N) cos 40 = 38.3 N Ans. F = 50 N F 40 3 F A = (50 N) sin 40 = N F A A

5 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 2, page 2 of 2 4 View of plane from above A 6 F BA = (32.14 N) sin 35 = 18.4 N F BA N 7 F = 18.4 N negative direction 35 F B 5 F = (32.14 N) cos 35 = 26.3 N Ans. 8 F = Fi + F j + Fk = {26.3i j 18.4k} N Ans.

6 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 3, page 1 of 2 3. Epress F in terms of,, and components A B F = 8 kn 1 View of the plane formed b A, F, and the ais. F A 3 F A = (8 kn) sin 70 = kn A 2 (8 kn) cos 70 = 2.74 kn 4 F = 2.74 kn Ans. F 70 Negative direction F = 8 kn

7 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 3, page 2 of 2 5 View of the plane from above. 6 F = (7.518 kn) cos 25 = 6.81 kn 7 F = (7.518 kn) sin 25 F F 25 B = 3.18 kn Ans kn 8 F = Fi + F j + Fk = {6.81i 2.74j k} kn Ans.

8 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 4, page 1 of 4 4. Determine the,, and components of the 26-N force shown. Also determine the coordinate direction angles of the force. F = 26 N A 20 B 1 View of the plane formed b A, F, and the ais. 2 F = 26 N F F = (26 N)( 5 ) 13 = 10 N Ans. A F A 3 F A = (26 N)( 12 ) 13 = 24 N

9 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 4, page 2 of 4 4 View of the plane as seen from above 7 Determine the coordinate direction angle,. (24 N) sin 20 = 8.21 N F 5 F = 8.21 N Ans., F = 26 N F = 8.21 N Negative direction 20 6 F = (24 N) cos 20 = 22.6 N Ans. F A A 24 N 8 View of the plane formed b the ais and the force F. B 8.21 N 9 The direction angle is measured from the the positive part of the ais. Here it is, not. 26 N 10 = 180 = 180 cos N 26 N = = Ans.

10 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 4, page 3 of 4 11 Determine the coordinate direction angle,. F = 26 N F = 10 N A B 12 View of the plane formed b the ais and the force F. 26 N 10 N 13 = cos N 26 N = 67.4 Ans.

11 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 4, page 4 of 4 14 Determine the coordinate direction angle,. 15 View of the plane formed b the ais and the force F. F = 26 N 26 N 22.6 N 16 = cos N 26 N = 29.6 Ans. A B F = 22.6 N 17 bservation: the calculations for,, and can be summaried b the general formulas cos = cos = F F F F cos = F F The algebraic signs of F, F, and F must be included when using these formulas.

12 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 5, page 1 of 3 5. Determine the magnitude and coordinate direction angles of the resultant of the three forces acting on the mast. 1 Epress F 1 in rectangular components. F 3 = 40 N 2 (100 N) sin 80 = N 30 F 2 = 60 N 3 F 1 = (98.48 N) cos 20 = N F 1 = 100 N F 1 = 100 N 5 4 F 1 = (100 N) cos 80 = N F 1 = (98.48 N) sin 20 = N 6 In vector form, F 1 = {33.68i 17.36j k} N (1)

13 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 5, page 2 of 3 7 Epress F 2 and F 3 in rectangular components. F 3 = 40 N 8 F 2 = (60 N) sin 30 = 30 N 10 F 3 = 40 N 30 F 2 = 60 N 9 F 2 = (60 N) cos 30 = N 11 In vector form, F 2 = {51.96i + 30j} N (2) F 3 = {40j} N (3) 12 Use Eqs. 1, 2, and 3 to compute the resultant, R. 13 Collect coeffiecients of i, j, and k. R = F 1 + F 2 + F 3 = {33.68i 17.36j k} N + {51.96i + 30j} N + {40j} N = { }i N + { }j N + {92.54k} N = {85.64i j k} N

14 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 5, page 3 of 3 14 Magnitude of resultant R = (85.64 N) 2 + (52.64 N) 2 + (92.54 N) 2 = N Ans. 15 Coordinate direction angles R cos = = R N N R cos = R = N N R cos = = R N N Solving for the angles gives = 51.2 Ans. = 67.4 Ans. = 47.4 Ans.

15 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 6, page 1 of 4 6. Determine the magnitude and coordinate direction angles of the resultant of the forces acting on the ee-bolt. F 1 = 650 N Epress F 1 in rectangular components. F 3 = 300 N F 2 = 800 N 2 F 1 = (650 N)( 5 13 ) = 250 N F 1 = 650 N (650 N)( 12 ) = 600 N F 1 = (600 N) cos 30 = N F 1 = (600 N) sin 30 = 300 N 6 In vector form, F 1 = { 519.6i + 250j 300k} N (1)

16 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 6, page 2 of 4 7 Epress F 2 in rectangular components. 10 Since we are not given 2, we cannot compute F 2 from the formula F 2 = F 2 cos 2 2 But since we know the magnitude F 2 = 800 N, we can solve for F 2 from the formula for the magnitude of a vector in terms of its rectangular components: F 2 = F F F 2 2 F 2 = 800 N Substituting for F 2, F 2, and F 2 gives (800 N) = (514.2 N) 2 + F (273.6 N) 2 and solving for F 2 gives F 2 = ±548.4 N 8 F 2 = (800 N) cos 50 = N Alternativel, we could have solved for 2 b using the identit satsified b the direction angles: 9 F 2 = (800 N) cos 70 = N (cos 50 ) 2 + (cos 2) 2 + (cos 70 ) 2 = 1 Solving gives 2 = and thus F 2 = (800 N) cos = N This is the same result as before, allowing for round-off error.

17 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 6, page 3 of 4 11 The figure shows that F 2 is positive, so choose the plus sign: F 2 = N F 2 12 In vector form, F 2 can be written as F 2 = {514.2i j k} N (2) F 2 13 Also, in vector form, F 3 is F 3 = {300k} N (3) F 3 = 300 N 14 Use Eqs. 1, 2, and 3, to compute the resultant R. R = F 1 + F 2 + F 3 = { 519.6i + 250j 300k} N + {514.2i j k} N + {300k} N = { }i N + { }j N + { }k N = { 5.4i j k} N

18 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 6, page 4 of 4 15 Magnitude of resultant R = ( 5.4 N) 2 + (798.4 N) 2 + (273.6 N) 2 = N Ans. 16 Coordinate direction angles R cos = R = 5.4 N N R cos = R = N N R cos = = R N N Solving gives = 90.4 Ans. = 18.9 Ans. = 71.1 Ans.

19 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 7, page 1 of 3 7. A 300-lb vertical force is required to pull the pipe out of the ground. Determine the magnitude and direction angles of the force F 2 which, when applied together with the 150-lb force F 1 shown, will produce a 300-lb vertical resultant. F 1 = 150 lb 45 1 Epress F 1 in rectangular components F F 1 = F 1 cos 1 = (150 lb) cos 60 = 75 lb F 1 = F 1 cos 1 = (150 lb) cos 60 = 75 lb F 1 = F 1 cos 1 = (150 lb) cos 135 = lb F 1 = 150 lb 1 = = = ( 1 measured from positive ) 2 In vector form, 1 = 60 F 1 = {75i + 75j 106.1k} lb (1)

20 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 7, page 2 of 3 3 Epress F 2 in terms of components, 4 Equating coefficients of i gives F 2 = F 2 i + F 2 j + F 2 k (2) Use Eqs. 1 and 2 to compute the resultant, R = F 1 + F 2 = {75i + 75j 106.1k} lb + {F 2 i + F 2 j + F 2 k} = {75 lb + F 2 }i + {75 lb + F 2 }j + { lb + F 2 }k (3) Now we use the fact that R is known to be vertical, directed upwards with a magnitude of 300 lb: R = 0i + 300j + 0k 0 = 75 lb + F 2 Solving gives F 2 = 75 lb (4) Equating coefficients of j, and then k, gives 300 = 75 lb + F 2 0 = lb + F 2 Solving gives F 2 = 225 lb (5) F 2 = lb (6) So Eq. 3 becomes 0i + 300j + 0k ={75 lb + F 2 }i + {75 lb + F 2 }j + { lb + F 2 }k

21 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 7, page 3 of 3 5 Magnitude of F 2 F 2 = (F 2 ) 2 + (F 2 ) 2 + (F 2 ) 2 = ( 75 lb) 2 + (225 lb) 2 + (106.1 lb) 2 = lb ns. Coordinate direction angles F 2 75 lb cos 2 = = lb F 2 F 2 cos 2 = = F lb lb Solving gives F 2 F 2 cos 2 = = lb lb F 2 2 = Ans. 2 2 = 30.0 Ans. 2 = 65.9 Ans. 2 2

22 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 8, page 1 of 3 8. Two forces, F 1 and F 2 act on the bracket as shown. If the resultant of F 1 and F 2 lies in the plane, determine the magnitude of F 2. Also determine the magnitude of the resultant. F 1 = 60 N F 2

23 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 8, page 2 of 3 1 Epress F 1 in rectangular components: F 1 = F 1 cos 1 = (60 N) cos 130 = N (1) F 1 = F 1 cos 1 = (60 N) cos 60 = 30 N (2) Since we do not know the value of 1, we cannot compute F 1 from F 1 F 1 = 60 N = 60 F 1 = F 1 cos 1 Instead we can use the equation for the magnitude of F 1 : F 1 = (F 1 ) 2 + (F 1 ) 2 + (F 1 ) 2 F F 1 1 = = 130 Solving gives 60 N = ( N) 2 + (30 N) 2 + (F 1 ) 2 F 1 = ± N 2 F 1 points in the negative direction so choose the minus sign F 1 = N (3) 3 In vector form, from Eqs. 1, 2, and 3, F 1 = { 38.57i + 30j 34.82k} N (4)

24 2.3 Rectangular Components in Three-Dimensional Force Sstems Eample 8, page 3 of 3 4 Now, Eq. 4 and the vector form of F 2, 5 Equating coefficients of k gives F 2 = F 2 k 0 = F N can be used to compute the resultant, R: or R = F 1 + F 2 = { 38.57i + 30j 34.82k} N + F 2 k (5) Because the resultant R lies in the - plane, F 2 = N Finall, the magnitude of R is R = (R) 2 + (R ) 2 + (R) 2 Ans. R = 0 (6) Combining Eqs. 5 and 6 gives = ( N) 2 + (30 N) 2 + (0) 2 = 48.9 N Ans. Ri + R j + 0k ={ 38.57i + 30j + (F )k} N Equating coefficients of i gives R = N Equating coefficients of j gives R = 30 N

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