7.5 Application of Vectors

Objectives:

(1) Solve inclined ramp problems.(2) Find the angle between two forces.

Example.Find the force required to pull a 50 lb weight up a 20◦ incline.

Solution.

Refer to the figure to the left.

Let−→BA represent the force of gravity.

Let−−→BC represent the force of the weight pushing against

the ramp.

Let −−→AC represent the force causing the weight to slide

down the ramp.

Let−−→BF represent the force required to prevent the weight

from sliding down the ramp.

It is not obvious, but, the magnitude of−−→BF and

−→CA is the same. Since the object is not

moving, the summation of the two forces is zero.

Since−−→BF is parallel to

−→CA and

−→BA is a transversal, ∠EBD ∼= ∠BAC. Since ∠BCA =

∠CBF = 90◦, 4ABC is similar to 4BED. This implies that ∠ABC = 20◦. Since 4ABCis a right triangle,

sin 20◦ =|−→AC

50⇒ |−→AC| = 50 sin 20◦ = 17.

Since−−→BF =

−→AC, the force required to pull the weight is 17 lb.

Example.A force of 25 lb is required to push an 80 lb weight up a hill. What angle does the hill makewith the horizontal?

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Solution.

Fortunately, we did all the hard work in the perviousexample. By similar reasoning, 4EBD is similar to4ABC. So, ∠ABC = θ.

sinB =25

80⇒ B = sin−1 25

80= 18◦

Example.Two forces of 692 Newtons and 433 Newtons act at a point. The resulting force is 786Newtons. Find the angle between the forces.

Solution.

We represent the forces withvectors and add them using theparallelogram law. The anglebetween the two forces is α +β.From the diagram we see thatthe side opposite α has length492. We shall use the Law ofCosines to find the measure ofα.

4232 = 6922 + 7862 − (2)(692)(786) cos α ⇒

cos α =6922 + 7862 − 4232

(2)(692)(786)

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α = 32.5◦

Now, use the Law of Cosines to solve for β.

6922 = 4232 + 7862 − (2)(423)(786) cos β ⇒

cos β =4232 + 7862 − 6922

(2)(423)(786)

β = 61.4◦

The angle between the tow forces is 93.9◦.

Example.A jet flies on a bearing of 233◦ at a speed of 450 mph. The wind is blowing at 39 mph froma direction of 114◦. Find the resulting bearing and ground speed of the plane.

Solution.

The vector with magnitude 450 rep-resents the air speed and directionof the plane. The vector with mag-nitude 39 represents the speed anddirection of the wind. The vec-tor labeled x represents the groundspeed and direction of the plane.

The angle between the plane vector and the north-south line is 233◦ − 180◦ = 53◦. The53◦ angle and the angle β are alternate interior angles formed by parallel lines cut by atransversal. Hence, β = 53◦. The angle between the wind vector and the north-south lineis 114◦. That angle, and the angle γ are alternate interior angles on the same side of atransversal. These angles are supplementary. So, γ = 180◦ − 114◦ = 66◦. So, the angle

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between the plane vector and the wind is 119◦. Now, we use the Law of Cosines to find themagnitude of x, which is the ground speed.

|x| =√

392 + 4502 − (2)(39)(450) cos 119◦ = 470

Use the Law of Sines to solve for α.sinα

39=

sin 119◦

470⇒ sinα =

39 sin 119◦

470⇒ α = 4.2◦

So, the resulting bearing is 237◦, and the ground speed is 470 mph.

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