Displacement vectors a, b, and c add up to a total of zero. vector a has a magnitude of 1550 m and a direction of 25.6° north of east. vector b has a direction of 41.0° east of south, and vector c has a direction of 35.1° north of west. find the magnitudes of vector b and vector c.

B= 5626.77 m

C= 6220. 5 m

Explanation:

Because the sum of the vectors must be equal to zero, then the result  force in x and the result force  in y must be  zero.

We propose 2 equations  x-y to solve the problem:

Rx :resulting from forces at x

Ry: resulting from forces at y

Rx= Ax+Bx+Cx=0

Ry= Ay+By+Cy=0

Ax =1550 *cos25.6°= 1397.84

Ay =1550 *sin25.6° = 669.73

Bx= B*sin41° = 0.656B

By= -B*cos41° = -0.7547 B

Cx= -C*cos35.1°= -0.8181 C

Cy= C* sin35.1° = 0.575 C

Rx= 1397.84+0.656B-0.8181 C=0

Ry= 669.73-0.7547 B+ 0.575 C=0

System of 2 equations with 2 incognites:

+0.656B-0.8181 C= - 1397.84

-0.7547 B+ 0.575 C= -669.73

Resolving the system:

B= 5626.77 m

C= 6220. 5 m

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

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