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Department of Mechanical Engineering, MITS-FET Name: Engineering Mechanics (ME-105) Tutorial. En. No: Spring Semester 2011-2012 Batch:

Engineering Mechanics (ME 105)

TUTORIAL SHEET-1

1. Determine the magnitude of the Resultant forces FR = F1 + F2 and its direction measured countercloockwise from the positive x axis.

Ans. FR = 867 N, =63.05o

2. Two forces are applied to an eye bolt fastened to a beam. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle law.

Ans: R=8.4 kN, 19o

3. Determine the resultant R of the two forces shown by (a) applying the parrologram law of vector addition (b) summing scaler component.

N, =79.1o

Ans. R= 529

4. Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive axis

5.

Determine the magnitude of the resultant force FR = F1 + F2 and its direction measured countercloockwise from the positive u axis.

Ans. FR =605 N,

= 85.4

o

6. The vertical force F acts downwards at A on the two membered frames. Determine the magnitudes of the two component of F directed along the axes of AB and AC. Set F = 500 N.

FAC = 366 N, FAB = 448 N

7. The plate is subjected to the two forces at A and B as shown. If = 600, determine the magnitude of the resultant of these two forces and its direction measured from the horizontal.

Ans: FR = 10.8 KN, = 33.16 , = 3.16 o0

8. To steady a sign as it is being lowered, two cables are attached to the sign at A. Knowing that = 250 determine a) the required magnitude of force P if the resultant R of the two forces applied at A is to be vertical, b) the corresponding magnitude of R

Ans: P = 108.6 N R =163.9 N

9. Determine the x and y component of each of the forces shown.

Ans: Fx1 = -320 N, Fy1 = 600 N, Fx2 = 360 N, Fy2 = 600 N, Fx3 = 600 N, Fy3 = -110 N

10.A window pole is used to open a window as shown. Knowing that the pole exerts on the window a force P directed along the pole and that the magnitude of the vertical component of P is 45 N, determine a) the magnitude of the force P, b) its horizontal component.

Ans. P= 47.9 N, Px = 16.38 N

11.Determine the magnitude and direction of F1 so that the resultant force is directed vertically upward and has a magnitude of 800 N.

Ans: F1 = 275 N, = 29.1

12.

Determine the resultant of the three forces of fig. shown.

Ans. R= 4.73 KN, 20.60

13. While steadily pushing the photocopying machine up an incline, a person exerts a 180 N force P as shown. Determine the components of P, which are parallel and perpendicular to the incline. Ans. Pt = 163.1 N, Pn = - 76.1 N

14.

Determine the resultant R of the two forces applied to the bracket.

Ans. Rx = 88.8 kN, Ry = 244.6 kN, Rx = 167.1 kN, Ry = 199.5 kN

15. It is desired to remove the spike from the timber by applying force along its horizontal axis. An obstruction A prevents direct access, so that two forces, one 1.6 kN and the other P, are applied by cables as shown. Compute the magnitude of P necessary to ensure axial tension T along the spike. Also find T. Ans. P = 2.15 kN, T = 3.20 kN

16. The guy cables AB and AC are attached to the top of the transmission tower. The tension in cable AC is 8 kN. Determine the required tension T in cable AB such that the net effect of the two cable tensions is a downward force at point A. Determine the magnitude R of this downward force. Ans. T = 5.68 kN, R = 10.21 kN