Question

Two forces, F1 and F2, act on a 1.00 kg object where F1 = 20.0 N and F2 = 12.0 N. 

(a) Find the acceleration in Figure P5.11a

(b) Find the acceleration in Figure P5.11 

Answer 1

Part A.

Net force is given by:

Fnet = F1 + F2 = m*a

Given that

F1 = (20.0 N) i

F2 = (12.0 N) j

m = 1.00 kg

So,

a = Fnet/m = (F1 + F2)/m

a = (20.0 i + 12.0 j)/1.00

a = 20.0 i + 12.0 j

Magnitude of acceleration = |a| = sqrt (20^2 + 12^2) = 23.3 m/sec^2

Direction = arctan (12/20) = 31.0 deg

Part B.

F1 = (20.0 N) i

F2 = 12.0 N at 60 deg with +axis

F2x = 12.0*cos 60 deg = (6.00 N) i

F2y = 12.0*sin 60 deg = (10.4 N) j

So,

Fnet = 20.0 i + (6.00 i + 10.4 j) = 26.0 i + 10.4 j

a = Fnet/m = Fnet/1.00 = Fnet

a = 26.0 i + 10.4 j

Magnitude of acceleration = |a| = sqrt (26.0^2 + 10.4^2) = 28.0 m/sec^2

Direction = arctan (10.4/26.0) = 21.8 deg

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