Problem 001 Problem Determine the x and y components of the forces shown below in Fig P-001. P-001.
Solution 001
Rectangular Representation
j. From the above vector notation of forces, Fx is the coefficient of i and Fy is the coefficient of j Problem 002 Compute the x and y components of each of the four forces shown in Fig. P-002. P-002.
Solution 002
Rectangular Representation
The coefficients of i and j from the vector notations are the respective x and y components of each force. Problem 003 Which of the following correctly defines the 500 N force that passes from A(4, 0, 3) to B(0, 6, 0)? A. 256i - 384 j + 192k N B. -256i + 384 j - 192k N C. -384i + 192 j - 256k N D. 384i - 192 j + 256k N
Solution 003
From the figure
Unit vector from A to B:
Rectangular representation of F:
Problem 004 Referring to Fig. 1.4, determine the angle between vector A and the y-axis. A. 65.7° B. 73.1° C. 67.5° D. 71.3°
Solution 004
Answer: D Problem 005 Find the components in the x, y, u and v directions of the force P = 10 kN shown in Fig. P-005.
Solution 005
Problem 006 The force P of magnitude 50 kN is acting at 215° from the x-axis. Find the components of P in u 157° from x, and v negative 69° from x.
Solution 006
From the figure:
Check: (okay!)
By Sine Law:
answer
answer Problem 007 A block is resting on an incline of slope 5:12 as shown in Fig. P-007. It is subjected to a force F = 500 N on a slope of 3:4. Determine the components of F parallel and perpendicular to the incline.
Solution 007
answer
answer Problem 008 A force P = 800 N is shown in Fig. P-008.
a. b. c. d.
Find the y-component of P with respect to x and y axis. Find the y'-component of P with respect to x' and y' axis. Find the y-component of P with respect to x' and y axis. Find the y'-component of P with respect to x and y' axis.
Solution 008
Part (a) y-component of P with respect to x and y axis
answer
Part (b) y'-component of P with respect to x' and y' axis
Part (c) y-component of P with respect to x' and y axis
The figure formed is in the shape of equilateral triangle. Thus,
answer
Part (d) y'-component of P with respect to x and y' axis
By Sine Law
answer Problem 009 The body on the 30° incline in Fig. P-009 is acted upon by a force P inclined at 20° with the horizontal. If P is resolved into components parallel and perpendicular to incline and the value of the parallel component is 1800 N, compute the value of the perpendicular component and that of P.
Solution 009
Perpendicular component
answer
Value of P
answer Problem 010 The triangular block shown in Fig. P-010 is subjected to the loads P = 1600 lb and F = 600 lb. If AB = 8 in. and BC = 6 in., resolve each load into components normal and tangential to AC.
Solution 010
Problem 016 The magnitude of vertical force F shown in Fig. P-016 is 8000 N. Resolve F into components parallel to the bars AB and AC.
Solution 016
By Sine Law:
answer
answer Problem 017 If the force F shown in Fig. P-017 is resolved into components parallel to th e bars AB and BC, the magnitude of the component parallel to bar BC is 4 kN. What are the magnitudes of F and its component parallel to AB?
Solution 017
By Sine law
answer
answer Problem 226 In Fig. P-226 assuming clockwise moments as positive, co mpute the moment of force F = 200 kg and force P = 165 kg about points A, B, C, and D.
Solution 226
Moment of force F about points A, B, C, and D:
→ answer
→ answer
→ answer
→ answer
Moment of force P about points A, B, C, and D:
(this means that point A is on the line of action of force P)
→ answer
→ answer
→ answer You can also resolve P to horizontal and vertical components at point E then take the moment of these components at point C. The answer would be the same. Try it.
→ answer Problem 227 Two forces P and Q pass through a point A which is 4 m to the right of and 3 m a moment center
O. Force P is 890 N directed up to the left at 30° with the horizontal and force Q is 445 N directed up to the left at 60° with the horizontal. Determine the moment of the resultant of these two forces with respect to O.
Solution 227
(to the right)
(upward)
(counterclockwise)
answer
The moment of resultant about O can be solved actually without the use of R x and R y. The moment effect of the components of R is the same as the combined moment effect of the components P and Q. Thus, . Try it. You can also find Mo by finding the magnitude of R and its moment arm about point O. Moment arm is the perpendicular distance between the line of action of R and point O. Problem 228 Without computing the magnitude of the resultant, c ompute where the resultant of the forces shown in Fig. P-228 intersects the x and y axes.
Solution 228
to the right upward
to the right downward
clockwise
to the right
upward
x-intercept of the resultant
to the left of point O
answer
y-intercept of the resultant
above point O
answer
Problem 229 In Fig. P-229, find the y-coordinate of point A so that the 361-lb force will have a clockwise moment of 400 ft-lb about O. Also determine the X and Y intercepts of the line of action of the force.
Solution 229
answer
Y-intercept of the line of action of force F
answer
X-intercept of the line of action of force F
answer Problem 230 For the truss shown in Fig. P-230, compute the perpendicular distance from E and from G to the line BD. Hint: Imagine a force F directed along BD and compute its moment in terms of its components about E and about G. Then equate these results to the definition of moment M = Fd to compute the required perpendicular distances.
Solution 230
Let d = length of member BD dx = 12 ft dy = 16 - 12 = 4 ft
Moment about point E
answer
Moment about point G
answer
Checking (by Geometry):
(okay!)
(okay!) Problem 231 A force P passing through points A and B in Fig. P-231 has a clockwise moment of 300 ft-lb about O. Compute the value of P.
Solution 231
Ratio and proportion
Moment at point O
down to the right from A to B
answer
Problem 232 In Fig. P-231, the moment of a certain force F is 180 ft·lb clockwise about O and 90 ft·lb counterclockwise about B. If its moment about A is zero, determine the force.
Solution 232
Moment about O
Moment about B
Substitute xFy = 180 to the above equation
Thus, F = 75 lb downward to the right at θx = 36.87° and x-intercept at (4, 0).
answer
Problem 233 In Fig. P-231, a force P intersects the X axis at 4 ft to the right of O. If its moment about A is 170 ft·lb counterclockwise and its moment about B is 40 ft·lb clockwise, determine its y intercept.
Solution 233
Resolve force P into components at its x-intercept
Resolve force P into components at its y-intercept
Thus, y intercept of force P is (0, -8/3).
answer
