Determine the magnitude of the resultant of the three coplanar forces in N. The

Question

Determine the magnitude of the resultant of the three coplanar forces in N. The magnitude of F1, F2 and F3 are 220, 280 and 400 N, respectively. The angles alpha, beta and gamma are 16, 51 and 172 degrees, respectively. A particle with a mass of 6.9 kg is at rest when the three forces shown in P2 suddenly act on it for 4.3 s. Determine the magnitude of the particle velocity in m/s. The magnitude of F1, F2 and F3 are 220, 280 and 400 N, respectively. The angles alpha, beta and gamma are 16, 51 and 172 degrees, respectively. Determine the direction of the velocity vector of the 6.9 kg particle in degrees when subjected to the forces shown in P2. The two tug boats pull the oil rig to travel in the direction shown in calm water. (See left directed arrow). Tug A pulls with a force of 70 kN directed at the angle alpha of 29 deg. and Tug B pulls with a force of 110 kN. In what direction must Tug B pull such that the oil rig travels in the direction shown? Determine the direction angle beta in deg. As shown in P5, Tug A pulls with a force of 70 kN directed at the angle alpha of 29 deg and Tug B pulls with a force of 110 kN and the oil rig travels in the direction shown at a constant velocity of 17 km/hr. Determine the 'drag' force exerted by the water on the oil rig in kN.

Explanation / Answer

(2)

The force F1

F1 = F1 cos alpha + F1 sin alpha

=220 cos16+220 sin 16

=211.47 i + 60.64 j

F2 = - F2 cos beta + F2 sin beta

= 280 cos(90+ 51) + 280 sin(90+ 51)

=-161.64 i + 130.89 j

F3 = F3 cos gama + F3 sin gamam

= 400 N cos (90+51 + 172 ) + 400 sin (90+ 51 + 172)

=272.79 i-292.54 j

net force

F = F1 + F2 + F3

= (211.47 -161.64 + 272.79 ) i +( 60.64 + 130.89 -292.54) j

=322.62 -101.01 j

magnitude of force is

F = sqrt ( 322.62 )^2 + (-101.01)^2

=338 N

As per guide lines I worked first problem kindly post remainiing questions in the next post

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