Set 5 Problem number 10

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s), Extension

• Figure(s)

Problem

You and I are pulling on a massive, initially stationary object resting on a smooth frozen pond.  You pull with a force of 2.51 pounds to the North and I pull with a force of 1.28 pounds to the East.  The object starts to move in response to our combined force.

• At what angle with the Easterly direction does the object initially move?
• How many pounds of force does the object experience from our combined pulls?

Solution

A sketch should consist of two vectors, one to the East with its 'tail' at the starting point and the other to the North with its 'tail' at the 'head' of the first.  

• These vectors form the sides of a right rectangle.

The equivalent force vector extends from the initial point along the hypotenuse, ending at the terminal point of the second vector.

• The legs of the triangle represent 1.28 pounds and 2.51 pounds, the direction of the hypotenuse represents the direction of the equivalent force, and the length of the hypotenuse represents the magnitude of the equivalent force.

• The length of the hypotenuse is found by the Pythagorean Theorem to be `sqrt( ( 2.51)² + ( 1.28)²) pounds = 2.82 pounds.
• The angle of the hypotenuse with East is tan^-1  ( 1.28 / 2.51) = 27.04 degrees.

Generalized Solution

We set up an x-y coordinate system with the x axis point East and the y axis pointing North.

• The force vectors to the East and to the North are represented tail-to-head in the x and y directions, respectively, starting from the origin of the coordinate system.
• The lengths of these vectors will be to a scale such that lengths represent the magnitudes of the forces.
• The terminal point of the second vector is at the other end of the hypotenuse from the starting point, and the equivalent force vector will be along the hypotenuse, with magnitude represented by the length of the hypotenuse.
• The length of the hypotenuse is

• length = `sqrt(Fx² + Fy²)

• The hypotenuse lies at angle

• angle = tan^-1(Fy / Fx)

• from the positive x axis; this is the angle the equivalent force makes with East.

Explanation in terms of Figure(s), Extension

The figure below shows the forces Fx and Fy as vectors parallel to the x and y axes.

• The path that results from starting at the origin and moving through displacement sx followed by displacement sy is indicated by the heavy (blue) lines forming the legs of the triangle.
• The equivalent, or resultant, force is the vector F represented by the arrow from the origin of the first vector to the terminal point of the second.

The magnitude of this force is represented by the length `sqrt(Fx² + Fy²) of the hypotenuse.

The angle of the displacement is tan^-1(Fy / Fx).

Figure(s)