MCQs

Equilibrium requires the additional force to be equal and opposite to the existing resultant. Therefore its components are -12 N and -5 N.

As the angle increases from 30° to 60° the cosine decreases while the sine increases. Therefore the x-component decreases and the y-component increases.

The resultant components are (R_x=4) N and (R_y=12) N. Therefore (tantheta=12/4=3), giving (thetaapprox71.6^circ), closest to 72°.

The resultant components are 5 m and 4 m. Thus (R=sqrt{5^2+4^2}=sqrt{41}) m.

(R_x=4+6-5=5) m and (R_y=3-3+4=4) m.

(V_y=sqrt{25^2-7^2}=24) N. Quadrant I requires a positive y-component.

The new perpendicular components are A and 2B. Hence (R=sqrt{A^2+4B^2}).

(V_y=sqrt{20^2-12^2}=16) N. In quadrant II the y-component is positive.

(V_x=Vcostheta). The x-component is zero when (costheta=0), which occurs at 90°.

(F=sqrt{9^2+12^2}=15) N.

Positive x and negative y components identify quadrant IV.

At 45° both components are equal. Each is (20cos45^circ=10sqrt2) N.

With only the 6 N vertical component remaining the magnitude is 6 N.

Both components cancel exactly. Therefore the resultant is zero.

The x-components cancel while the y-components add to 8 N upward.

If both components of the resultant are zero then the resultant magnitude is zero.

(R_x=0) and (R_y=12) N. Therefore the resultant is 12 N upward.

The components are perpendicular so their squares add to the square of the resultant magnitude.

The first force contributes (100sin30^circ=50) N upward. Adding the 50 N vertical force gives 100 N upward.