Problem 001Problem Determine the x and y components of the forces shown below in Fig P-001. 

 

Solution 001

 

 

 

 

Rectangular Representation

 

From the above vector notation of forces, Fx is the coefficient of i and Fy is the coefficient of j.

Problem 002Compute the x and y components of each of the four forces shown in Fig. 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 003Which of the following correctly defines the 500 N force that passes from A(4, 0, 3) to B(0, 6, 0)?A. 256i - 384j + 192k NB. -256i + 384j - 192k NC. -384i + 192j - 256k ND. 384i - 192j + 256k N 

 

Solution 003

From the figure

 

Unit vector from A to B:

 

Rectangular representation of F:

Problem 004Referring 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 005Find the components in the x, y, u and v directions of the force P = 10 kN shown in Fig. P-005. 

 

Solution 005

 

 

Problem 006The 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 007A 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 008A force P = 800 N is shown in Fig. P-008.

a. Find the y-component of P with respect to x and y axis.b. Find the y'-component of P with respect to x' and y' axis.c. Find the y-component of P with respect to x' and y axis.d. 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 009The 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 010The 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 016The 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 017If the force F shown in Fig. P-017 is resolved into components parallel to the 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 226In Fig. P-226 assuming clockwise moments as positive, compute 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 227Two 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 Rx and Ry. 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 228Without computing the magnitude of the resultant, compute 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 229In 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 230For 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

Letd = length of member BDdx = 12 ftdy = 16 - 12 = 4 ft

 

 

Moment about point E

 

          answer 

Moment about point G

 

          answer 

Checking (by Geometry): 

 

 

          (okay!) 

          (okay!)

Problem 231A 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 232In 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 233In 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