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 39 lb. Johnny pulls at an angle of 44° south of due west with a force of 68 lb. Connie pulls with a force to be determined.

Question

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 39 lb. Johnny pulls at an angle of 44° south of due west with a force of 68 lb. Connie pulls with a force to be determined.

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

Paul:

2)

3) Johnny:

4)

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

Connie:

6)

7) Magnitude:

8) Direction

Thank you!

Answer 1

Paul: F1 = 39Lb[45o] = 27.6 + 27.6i.

Johnny: F2 = 68Lb[224o] = -48.9 - 47.2i.

F1+F2 = (27.6+27.6i) + (-48.9-47.2i),
F1+F2 = -21.3 - 19.6i = 28.9Lb[42.6o]S. of W. = 28.9Lb[222.6] CCW.

Connie:
Connie's force is the equilibrant force, because it reduces the resultant force to zero:

F3 = -(F1+F2) = -(-21.3-19.6i) = 21.3 + 19.6i = 28.9Lb[42.6o] N. of E. = 28.9Lb[42.6o] CCW.

Answer 2

1) To express the boys' forces in terms of unit vectors, we can use the following convention:

- The positive x-direction is to the right.
- The positive y-direction is upwards.

Paul pulls to the northeast, which means his force has components in both the x and y directions. We can break down his force as follows:

Paul's force = 39 lb (cos 45° î + sin 45° ĵ)
= 39 lb (√2/2 î + √2/2 ĵ)
= 19.5√2 lb î + 19.5√2 lb ĵ

Johnny pulls at an angle of 44° south of due west. We can break down his force as follows:

Johnny's force = 68 lb (cos (-44°) î + sin (-44°) ĵ)
= 68 lb (cos 44° î - sin 44° ĵ)

2) Now let's calculate the components for Johnny's force:

Johnny's force = 68 lb (cos 44° î - sin 44° ĵ)
= 68 lb (0.7193 î - 0.6947 ĵ)

3) Now let's determine the third force from Connie. Since the sled does not move, the net force acting on it must be zero. This means the vector sum of all the forces must be zero.

To find Connie's force, we need to use vector addition. We can write the equation:

Paul's force + Johnny's force + Connie's force = 0

Since we already know the components of Paul's and Johnny's forces, we can substitute them into the equation:

(19.5√2 lb î + 19.5√2 lb ĵ) + (0.7193 lb î - 0.6947 lb ĵ) + Connie's force = 0

4) Rearranging the equation, we can isolate Connie's force:

Connie's force = -(19.5√2 lb î + 19.5√2 lb ĵ) - (0.7193 lb î - 0.6947 lb ĵ)

Connie's force = -19.5√2 lb î - 19.5√2 lb ĵ - 0.7193 lb î + 0.6947 lb ĵ

Connie's force = (-19.5√2 - 0.7193) lb î + (-19.5√2 + 0.6947) lb ĵ

Finally, simplify:

Connie's force = (-19.5√2 - 0.7193) lb î + (-19.5√2 + 0.6947) lb ĵ

5) The components of Connie's force are (-19.5√2 - 0.7193) lb in the x-direction and (-19.5√2 + 0.6947) lb in the y-direction.

To find the magnitude and direction of Connie's force, we can use:

Magnitude (|F|) = √[(x-component)^2 + (y-component)^2]

Direction (θ) = arctan(y-component / x-component)

Magnitude:

|Connie's force| = √[(-19.5√2 - 0.7193)^2 + (-19.5√2 + 0.6947)^2] lb

Direction:

θ = arctan[(-19.5√2 + 0.6947) / (-19.5√2 - 0.7193)] radians (convert to degrees if desired)

This will give you the magnitude and direction (angle) of Connie's force.