Question

An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive x direction and has a magnitude of 5.1 N; a second force has a magnitude of 4.3 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

Answer 1

The sum of the three force vectors must be zero since there is no acceleration.

The third force will be the equilibrant (which is minus the resultant) of the first two.

Since the first two forces are perpendicular, the equilibrant magnitude is given by the Pythagorean theorem. It is 6.67 N

Answer 2

When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop?

Answer 3

F1+F2+F3 = M*a = M*0 = 0.
5.1 - 4.3i + F3 = 0.
F3 = -5.1 + 4.3i = 6.67[40.1o] N. of W. = 139.9o CCW from +x-axis.

Answer 4

Wb = M*g = 1.70 * 9.8 = 16.66 N. = Normal force(Fn).

Fap-Fs = M*a = M*0 = 0.
2.6 - u*Fn = 0.
2.6 - u*16.66 = 0.
16.66u = 2.6.
u = 0.156 = Coefficient of static friction.

1.5-Fk = M*a = M*0 = 0.
1.5 - u*Fn = 0.
1.5 - u*16.66 = 0.
16.66u = 1.5.
u = 0.090 = Coefficient of kinetic friction.