Determine the magnitude of the resultant force fr=f1+f2

Answer

General guidance

Concepts and reason
The concepts required to solve this problem is the resolution of vectors.

Initially, find the horizontal and the vertical component of the force vector and represent it in vector form. Then, find the horizontal and the vertical component of the force vector and represent it in vector form.

Later, add both the forces in the vector form. Add the horizontal components of both the force vectors together and the vertical components of both the force vector together.

Finally, calculate the magnitude and the angle of the resultant force using the horizontal and the vertical components of the force.

Fundamentals

A vector F can be represented as:

F3FX+FЎ

Here, F is the x-component of the force, is the y-component of the force, and are the unit vectors which show the direction of the force in the x-axis and the y-axis respectively.

The expression for the magnitude of the resultant force is,

Е+F.

Here, F is the x-component of the force and is the y-component of the force.

The expression of the direction of the resultant force is,

p= tan
F

Here, is the angle between the x-component of the force F and the y-component of the force .

Step-by-step

Step 1 of 4

(a)

Refer the figure given in the question.

The horizontal and the vertical components of the force vector is,

Fsin0
co

Here, is the horizontal component of the force , is the vertical component of the force , is the angle of the force making with the positive y-axis.

Substitute 250 lb for and 30° for .

F -(250 lb)sin 30°
125 lb
Fy(250 lb) cos 30°
=216.50 lb

The vector form of the force is,

Substitute 125 lb for and 216.50 lb for .

(125 lb)+(216.50 lb)^

The horizontal component of the force vector is in positive x-direction. The vertical component of the force vector is in positive y-direction.

Step 2 of 4

Refer the figure given in the question.

The horizontal and the vertical components of the force vector is,

cos
=-F, sin^
2y

Here, is the horizontal component of the force , 2y is the vertical component of the force , is the angle of the force making with the positive x-axis.

Substitute 375 lb for and 45 for .

F2 (375 lb) cos 45°
265.17 lb
Fy =-(375 lb)sin 45°
=-265.17 lb

The vector form of the force is,

2х

Substitute 265.17 lb for and -265.17 lb for 2y.

F (265.17 lb)-(265.17 Ib)

The horizontal component of the force vector is in positive x-direction. The vertical component of the force vector is in negative y-direction. The force is equal to the vector sum of the horizontal and the vertical component of the force .

Step 3 of 4

The resultant force using the vector addition of the force and is,

FRFF

Substitute (125 lb)f+(216.50 lb) for and (265.17 lb)î-(265.17 lb)ỹ for .

F(125 lb)+(216.50 Ib)^+(265.17 lb)f-(265.17 lb)^
=(125 lb+265.17 lb)f+(216.50 lb-265.17 lb)
(390.17 lb)f-(48.67 lb)

The expression for the magnitude of the resultant force is,

2
F
R

Substitute 390.17 lb for F and -(48.67 lb) for .

||F= (390.17 Ib) +(-48.67 lb)
= 393.2 lb

Part a

The magnitude of the resultant force is 393 lb.

The resultant force can be calculated by the vector addition of the forces and . The resultant vector is the sum of the two vectors. The expression for the resultant vector is,

FRFF

Here, are the two vectors.

Step 4 of 4

(b)

The expression of the direction of the resultant force FR is,

p= tan
F

Substitute 390.17 lb for F and -(48.67 lb) for .

-48.67 lb
P= tan
390.17 lb
=-7.11°

The angle measured in counterclockwise direction from the x-axis is,

p360-p

Substitute 7.11° for .

p=360°-7.11°
= 353°

Part b

The direction of the resultant force is 353°.

The direction of the resultant force can be calculated using the horizontal component and the vertical component of the resultant force. The angle is measured in the clockwise direction from the positive x-axis. To measure the angle in the counter-clockwise direction, the angle is subtracted from to measure the angle in counterclockwise direction from the x-axis.

Answer

Part a

The magnitude of the resultant force is 393 lb.

Part b

The direction of the resultant force is 353°.

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