A differential equation encountered in the vibration of beams is
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International Edition---engineering Mechanics: Statics, 4th Edition
- The distance s (in m) above the ground for a projectile fired vertically upward with a velocity of 4 [1] m/s as a function of time t (in s) is given by s=4 [1]t-4.[8]t2 Find t for v = 0; find v for t = 4 and find v for t = 5arrow_forwardA particle is subjected to the action of two forces: F1 = 4 i+7j+7 k [kN] and F2 = 5 i - 6 j - 5 k [kN]. The coordinate direction angle of the resultant force with X-axis is: Select one: a. ar = 83.81° b. ar = 31.93° C. Ar = 13.93° O d. ar = 166.07°arrow_forwardVectors A, Band C are defined as follows: A = 7.5i + 7j - 4.3k; B = 2.5i – 8.3j + 7k; C = 4.8i + 2.9k. Determine the magnitude of vector A [units]. Answer: Determine the magnitude of vector B [units]. Answer: Determine the dot product of vectors A and B. Answer: Determine the angle between vectors A and B in degrees. Answer: Determine the magnitude of the vector (cross) product between vectors A and C. Answer: Determine AxB.C Answer: Determine A (BxC) Answer:arrow_forward
- If a vector force F = 6 i-9 j +2 k [kN], what will be the magnitude of this force: Select one: a. F=11 [kN] Ob. F=12 [kN] O c. F=0 O d. F=-11 [kN] metere to search IIarrow_forwardA force-couple system is acting on the frame as shown. Use A-50N, B=500N, C=80N, and M= 50N.m. The system is to be replaced with a single resultant force R. B [Select] [ Select] [Select] [Select] degrees. [ Select] 30° action in mm. [Select] action in mm. M 400 mm Y Y 200 mm Y 300 mm (a) Calculate the absolute value of the x-component of the resultant in N. (b) Calculate the absolute value of the y-component of the resultant in N. ▼ (c) Calculate the magnitude of resultant in N. (d) Calculate the angle that the resultant makes with the horizontal in ▼ (e) Calculate the absolute value of the x-intercept of the resultant's line of Y (f) Calculate the absolute value of the y-intercept of the resultant's line ofarrow_forward5. Forces [35k], [-35k] and [50i] , [-50i] are shown in the figure. Distances in the figure: 250 mm, 350 mm, |AB|=d mm, the angle between AB line and xy plane is 30°. (35k) N A) vector product of CA vector and 50 i (force) vector 250 mm B) vector product of AB vector and 50 i (force) vector |-50i) N (-35k) N C) scaler product of AC vector and 50 i (force) vector 350 mm D) vector product of CB vector and 50 i (force) vector (S0i) N E) scaler product of BA vector and 50 i (force) vector What is the couple moment of 50i and -50i forces?arrow_forward
- F1=4i+3j+6 k [kN] and F2 = 5 i-6j-5k [kN]. The coordinate direction angle of %3D %3D the resultant force with the X-axis is:arrow_forward6. If V(x,y) = xsin x cos y + y² then, F = -VV and it is equal to A. - ily sin x cos y-sin x cos y] +j[-xsin x sin y + 2x] B. -ix cos x cos y -cos x cos yl-j[-xsin x sin y + 2y] C. - ily cos x siny - sin x cos y] + [-xsin x sin y + 2y] D. - [[x cos x cos y + sin x sin y] - [-xsin x siny + 2y] E. -lly sin x cos y-sin x cos y] + [-xsin x siny + 2y] F. - [[x cos x sin y-sin x cos y]-[-xsin x siny + 2y] G. -[x cos x cos y + sin x cos y]-[-xsin x siny + 2y] H. None of the above A OB O C O D E LL G O Harrow_forwardAs indicated, three blocks were suspended in the air by massless strings that passed through massless and fixed pulleys. m (middle mass) = 5.00 [kg]. When the system reached equilibrium, it was discovered that a = 60.0° and y = 30.0 Draw the free body diagrams of block A, B, and C. MA 1. Write down Newton's 2nd law along the x and y directions for all three blocks. You must have six equations total. 2. What is the magnitude of mA? 3. What is the magnitude of mc? 4. What would be the magnitude of mA and mc when a = y? MB mcarrow_forward
- (The complete question is in the picture) The force acting on an object is F = 3.50 [N]ˆi−5.20 [N]ˆj−2.30 [N]kˆ. The vector from the origin to the point where the force is applied is r = 3.50 [m]ˆi + 2.50 [N]ˆj. What is the vector torque with respect to the origin produced by this force?A. 10.9 [N.m]ˆi + 11.5 [N.m]ˆj − 27.0 [N.m]kˆB. −18.2 [N.m]ˆi + 8.75 [N.m]ˆjC. −5.75 [N.m]ˆi + 8.05 [N.m]ˆj − 27.0 [N.m]kˆD. −5.75 [N.m]ˆi − 8.05 [N.m]ˆj + 9.45 [N.m]karrow_forwardsecond moments=475.15 Determine, using the method of composite areas, the second moments of area of part 2 about the y-axis [units^4]. Determine, using the method of composite areas, the second moments of area of part 3 about the y-axis [units^4]. Determine, using the method of composite areas, the total second moments of area of of the composite about the y-axis [units^4].arrow_forward3 m 3 m 2 m E 6 m 4 m F - {350j) N F = {-800k} N Figure 4 [Rajah 4]arrow_forward
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