Question

Three forces are acting on a stationary object. Force A is 360N acting at 64 degrees North of East. Force B is 280N acting at 40 degrees West of North and Force C of 420N acting at -150 degrees. If the mass of the object is 100 kg and the coefficient of friction is 0.44 determine the velocity of the object after 8 seconds (its response requires a direction).

Answer 1

The combined action of the 3 forces results in a total force where F₁ is the force of 360 N acting at an angle of 64⁰ from the +ve x-axis, F₂ is the force of 280 N acting at an angle of 50⁰ from the -ve x-axis and F₃ is the force of 420 N acting at an angle of 30⁰ from the -ve x-axis.

Thus the x-component of the combined force F_x=360cos64⁰-280cos50⁰-420cos30⁰=-386 N and the y-component of the combined force F_y=360sin64⁰+280sin50⁰-420sin30⁰=324+214-210=328 N.

The resultant force has a magnitude and a direction such that is the angle the force makes with the -ve x-axis.

Since the coefficient of friction is the frictional force acting on the object=F_f= so that the net force on the body due to all the forces= in a direction 40⁰ with the -ve x-axis.

Since , Therefore the velocity of the body after 8 secs=

.