A sled at rest is suddenly pulled in three horizontal directions at the same time, but it does not move. Paul pulls to the northeast with a force of 31 lb. Johnny pulls at an angle of 36° south of due west with a force of 52 lb. Connie pulls with a force to be determined.

1) Express the boys' two forces in terms of the usual unit vectors

2)Determine the third force (from Connie), expressing it first in component form and then as a magnitude and direction (angle). Neglect friction.

component of paul force:31cos45i+31sin45j=21.9i+21.9j component of johny force:52cos216i+52sin216j=-42.1i-30.6j Connie force is the resultant=21.9i+21.9j-42.1i-30.6j=-20.2i-8.7j direction=arctan(-8.7/-20.2)=23.3degree Fmag=sqrt(20.2^2+8.7^2)=21.99N

1) Expressing the boys' two forces in terms of the usual unit vectors:

Let's use the following unit vectors:

i: Represents the positive x-axis direction (east)
j: Represents the positive y-axis direction (north)

a) Paul's force:
Given that Paul pulls to the northeast, which can be represented as a vector pointing in the direction of the positive x-axis and positive y-axis, we can break down his force as follows:

Paul's force = 31 lb × (cos(45°)i + sin(45°)j) = 31 lb × (√2/2 i + √2/2 j) = 31√2/2 lb (i + j)

b) Johnny's force:
Johnny's force is at an angle of 36° south of due west. This means his force points in the negative x-axis direction and 36° below the negative x-axis direction. We can break down his force as follows:

Johnny's force = 52 lb × (-cos(306°)i + sin(306°)j) = 52 lb × (-1/2 i - √3/2 j) = -26 lb (i + √3/2 j)

2) Determine Connie's force:

Since the sled does not move, the net force acting on it in the x-axis and y-axis directions must be zero. Let's consider the x-axis first:

In the x-axis direction, the forces acting on the sled are Paul's force and Johnny's force. The x-axis components of these forces should cancel each other out:

Paul's x-component = 31√2/2 lb × i
Johnny's x-component = -26 lb × i

For the x-components to cancel, we must have:

31√2/2 lb × i + (-26 lb × i) = 0

Simplifying this equation gives us:

31√2/2 lb = 26 lb

Now, let's consider the y-axis:

In the y-axis direction, the forces acting on the sled are Paul's force, Johnny's force, and Connie's force. The y-axis components of these forces should cancel each other out:

Paul's y-component = 31√2/2 lb × j
Johnny's y-component = √3/2 × (-26 lb) × j
Connie's y-component = Fy lb × j (where Fy represents the y-component of Connie's force)

For the y-components to cancel, we must have:

31√2/2 lb + √3/2 × (-26 lb) + Fy lb = 0

Simplifying this equation gives us:

Fy lb = √3/2 × 26 lb - 31√2/2 lb

To express Connie's force as a magnitude and direction (angle), we can combine the x-axis and y-axis components. The resultant force, R, can be calculated as follows:

R^2 = (31√2/2 lb)^2 + (Fy lb)^2

Simplifying this equation gives us:

R^2 = 31^2/2 lb^2 + (√3 × 26 lb - 31√2/2 lb)^2

Once we calculate the value of R, we can determine its direction (angle) by using the formula:

θ = arctan(Fy lb / (31√2/2 lb))

Please note that without knowing the values of √3 and √2, we can only provide an algebraic expression for the magnitude and direction of Connie's force.

To solve this problem, we will break down the forces into their respective components and then add them up to find the net force on the sled. Remember that forces are vector quantities, so they have both magnitude and direction.

1) To express the boys' two forces in terms of the usual unit vectors, let's assign the positive x-axis to be due east and the positive y-axis to be due north. The unit vectors in these directions are î (east) and ĵ (north).

Paul's force, which is pulling in the northeast direction, can be resolved into its x and y components. The northeast direction can be divided into equal components of east and north. Since they are equal, the components have the same magnitudes.

The magnitude of Paul's force is 31 lb. So, the x-component is 31 lb multiplied by cos(45°) since the east direction is at a 45° angle with the positive x-axis. Therefore, the x-component of Paul's force is 31 lb * cos(45°) * î.

And the y-component is 31 lb multiplied by sin(45°) because the north direction is at a 45° angle with the positive y-axis. So, the y-component of Paul's force is 31 lb * sin(45°) * ĵ.

Johnny's force, which is pulling at an angle of 36° south of due west, also needs to be resolved into its x and y components. Firstly, convert the angle to its counter-clockwise angle from the positive x-axis.

We have an angle of 36° south of due west, which is equivalent to 180° + 36° = 216°. Now, subtract this angle from 360° to find the clockwise angle from the positive x-axis.

The clockwise angle is 360° - 216° = 144°. Therefore, the angle with the positive x-axis is 144°.

The magnitude of Johnny's force is 52 lb. So, the x-component is 52 lb multiplied by cos(144°) since the cos function gives the x-component for angles measured from the positive x-axis. Thus, the x-component of Johnny's force is 52 lb * cos(144°) * î.

And the y-component is 52 lb multiplied by sin(144°) since the sin function gives the y-component for angles measured from the positive x-axis. Hence, the y-component of Johnny's force is 52 lb * sin(144°) * ĵ.

2) To determine the third force from Connie, we will add up the x-components and y-components of the forces to find the net force acting on the sled.

Adding up the x-components:
Net x-force = Paul's x-component + Johnny's x-component = 31 lb * cos(45°) * î + 52 lb * cos(144°) * î.

Adding up the y-components:
Net y-force = Paul's y-component + Johnny's y-component = 31 lb * sin(45°) * ĵ + 52 lb * sin(144°) * ĵ.

To express the net force in component form, combine like terms:
Net force = (31 lb * cos(45°) + 52 lb * cos(144°)) * î + (31 lb * sin(45°) + 52 lb * sin(144°)) * ĵ.

To find the magnitude and direction of the net force, we can use the Pythagorean theorem and the inverse tangent function:

Magnitude of net force = sqrt[(Net x-force)^2 + (Net y-force)^2]

Angle with the positive x-axis = tan^(-1)(Net y-force / Net x-force)

Substituting the values we found earlier:
Magnitude of net force = sqrt[((31 lb * cos(45°) + 52 lb * cos(144°))^2 + ((31 lb * sin(45°) + 52 lb * sin(144°))^2]

Angle with the positive x-axis = tan^(-1)([31 lb * sin(45°) + 52 lb * sin(144°)] / [31 lb * cos(45°) + 52 lb * cos(144°)])

Calculating these values will give us the magnitude and direction of the net force (Connie's force) acting on the sled.