A 6.00 mass has three forces acting on it as shown in the table.

a. Determine the net force on the mass (in N).

b. Dtermine the net acceleration of the mass(in m/s^2)

Table:

Fa 100.0N due North

Fb 80.0N at 40.0 degree SofW

Fc 110N at 20.0 degree SofE

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The object with mass 6.00 kg mass has three forces acting on it, Fa 100.0 N due North, Fb 80.0 N at 40.0 degree S of W, Fc 110N at 20.0 degree S of E.

Take the components of each of these forces in the South-North directions. This gives 100 N due North, 80*sin(40) = 51.42 due South and 110*sin 20 = 37.22 N due South. The net of these is force is 11.36 due North. Take the components of the forces in the East-West directions. The net force is 110*cos 20 - 80*cos 40 = 42.08 N due East.

Combining the components due East and due North gives the magnitude of the net force as 43.58 N in the direction is 15.18 degrees North of East.

The acceleration due to this force is 43.58/6 = 7.26 m/s^2 due 15.18 degrees North of East.

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