Question

Forces A and B have magnitudes of 10.0 N and 20.0 N, respectively. What is the magnitude of their sum when A is directed along \theta=0^\circθ = 0 ∘ and B is directed along \theta=50^\circθ = 50 ∘?

Answer 1

Answer:

The magnitude of the sum of A and B is 27.51 N.

Explanation:

We can separate the x- and y-components of the vectors, then sum these components separately. Afterwards, we can calculate the magnitude of the sum.

$\vec{A}_x = A\cos{\theta} = A\cos{0^\circ} = 10~N\\\vec{A}_y = A\sin{\theta} = 0$

On the other hand, for the vector B:

$\vec{B}_x = B\cos{theta} = B\cos{50^\circ} = 20(0.6428) = 12.85~N\\\vec{B}_y = B\sin{50^\circ} = 20(0.766) = 15.32~N$

The sum of the components in the x-direction is equal to 22.85 N.

The sum of the components in the y-direction is equal to 15.32 N.

Finally, the magnitude of the sum vector C is

$C = \sqrt{(22.85)^2 + (15.32)^2} = 27.51~N$

Answer 2

Answer:27.51 N

Explanation:A=Magnitude of Force A= 10 .0 N

B=Magnitude of Force B=20.0 N

Direction of force A with horizontal = 00

Direction of force B with horizontal= 500

Ax =X component of force A= A cos (0) =10 cos (0) = 10.0 N

Bx =X component of force B=B cos (0) = 20 cos (50) =12.85 N

Ay =Y component of force A= A sin (50) = 10 sin (0) = 0 N

By =Y component of force B= B sin (50) = 20 sin (50) = 15.32 N

Rx=X component of resultant R= Ax + Bx = 10+ 12.85= 22.85 N

Ry=Y component of resultant R=Ay + By = 0+ 15.32 =15.32N

Magnitude of resultant R= sqrt(Rx2+ Ry2) = sqrt(22.852+ 15.322)=27.51 N

Direction of resultant=TAN-1 = TAN-1= 33.840