Question

You are told that an object has zero acceleration. Two forces on the object are: F_1 has magnitude 2.4 N at 45 degrees above the +x-axis and F_2 has magnitude 4.8 N at 37 degrees below the -x-axis. What is the third force that must be acting on the object? THE ANSWER ALREADY PROVIDED IS CONFUSING, I THINK

A) The vector F_3 has an x-component 5.5 N in the -x-direction and a y-component 4.6 N in the -y-direction.

B) The vector F_3 has an x-component 5.5 N in the +x-direction and a y-component 4.6 N in the +y-direction.

C) The vector F_3 has an x-component 3.8 N in the -x-direction and a y-component 2.9 N in the -y-direction.

D) The vector F_3 has an x-component 2.1 N in the -x-direction and a y-component 1.2 N in the -y-direction.

E) The vector F_3 has an x-component 2.1 N in the +x-direction and a y-component 1.2 N in the +y-direction.

Answer 1

net acceleration is xzero it means the vector sum of F1, F2and F3 is zero

so

F1 + F2 + F3 = 0

2.4 ( cos 45 i + sin 45 j) + 4.8* ( - cos 37 i - sin 37 j) + F3 = 0

F3 = + 2.136 i + 1.19 j

so option (d) is correct

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