Question

Only two horizontal forces act on a 1.2 kg body. One force is 6.4 N, acting due east, and the other is 7.4 N, acting 64° north of west. What is the magnitude of the body's acceleration?

Answer 1

Using east as +x and north as +y.

Note that the direction of the 7.4N force is 64⁰ north of west, which means 26⁰ anticlockwise from north.

x component of 6.4N force = 6.4N
y component of 6.4N force = 0

x component of 7.4N force = -7.4cos(64⁰) = -3.24N
y component of 7.4N force = 7.4sin(64⁰) = 6.65N

Total x component = 6.4 + (-3.24) = 3.16N
Total y component = 0 + 6.65 = 6.65N

Magnitude of force = √(3.16² + 6.65²) = 7.36N

Since F=ma, a = F/m
a = 7.36/1.2 = 6.14 m/s²