a) Determine the direction of the velocity vector of the 6.4 kg particle in degr

Question

a) Determine the direction of the velocity vector of the 6.4 kg particle in degrees when subjected to the forces shown in P2. Note: ‘Direction’ is the angle from the x-axis to the velocity vector. The magnitudes of F1, F2 and F3 are 400, 340, and 400 N, respectively. The angles alpha, beta and gamma are 35, 55 and 185 degrees, respectively.

b) The center of mass of the 45.1 kg chandelier shown in P7 hangs on a chain a distance L = 2.4 m to a hook in the ceiling. While traveling on level tracks, the train slows at a constant rate of 40400 km/hr/hr. Determine the tension in the chain during this deceleration in N. Note the units for acceleration, km/hr/hr, should be converted to the usual m/s/s.

As shown above, Tug A pulls with a force of 95 kN directed at the angle alpha of 35 deg and Tug B pulls with a force of 145 kN and the oil rig travels in the direction shown at a constant velocity of 15 km/hr. Determine the 'drag' force exerted by the water on the oil rig in kN.

Explanation / Answer

let i and j are unit vectors in the direction of x and y axis respectively.

as the oil rig travels with constant velocity , net force on it are zero.

force A=95*(cos(35) i + sin(35) j) kN

force B=145*(cos(beta) i- sin(beta) j)

let drag force magnitude be F. it is acting along -ve x axis.

then adding the three forces and equating x component and y component to 0:

along y axis:

95*sin(35)-145*sin(beta)=0

==>beta=22.073 degrees

along x axis:

95*cos(35)+145*cos(beta)-F=0

==>F=212.19 kN

hence magnitude of drag force is 212.19 kN

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