Question

A force of 8 N makes an angle of π/4 radian with the y-axis, pointing to the right. The force acts against the movement of an object along the straight line connecting (1, 3) to (4, 5).

a. Find a formula for the force vector F

b. Find the angle θ between the displacement direction D = (4 − 1)i + (5 − 3)j and the force direction F. (Round your answer to one decimal place.)

c. The work done is F · D or, equivalently, ||F||||D||cos(θ). Compute the work from both formulas and compare

Answer 1

To find the formula for the force vector F, break the force vector into its x and y components using trigonometry.

Step-by-step explanation:

To find the formula for the force vector F, we can use the given information that the force has a magnitude of 8 N and makes an angle of π/4 radians with the y-axis. We can break the force vector into its x and y components using trigonometry. The x-component of the force can be found by multiplying the magnitude of the force by the cosine of the angle, and the y-component can be found by multiplying the magnitude of the force by the sine of the angle.

b. To find the angle θ between the displacement direction D and the force direction F, we can use the dot product formula: cos(θ) = (F * D) / (|F| * |D|). This will give us the cosine of the angle, and we can take the inverse cosine to find the angle.

c. To compute the work done, we can use the formula W = F * D, which is equivalent to W = |F| * |D| * cos(θ), where |F| is the magnitude of the force, |D| is the magnitude of the displacement, and θ is the angle between the force and displacement.

Answer 2

a) F= 4
`√(2)` i + 4
`√(2)` j

b) Θ=11.3

c) The work done is 20
`√(2)`

Step-by-step explanation:

a) ||F||=8, α=π/4

Fx=||F||·sin(π/4)=8·
`(√(2))/(2)`

Fy=||F||·cos(π/4)=8·
`(√(2))/(2)`

F=Fx i + Fy j = 4
`√(2)` i + 4
`√(2)` j

b) We can find the value of Θ using the equation:

cos(Θ)=
`(F.D)/(||F||||D||)`

where:

D= 3 i + 2 j

F=4
`√(2)` i + 4
`√(2)` j

The dot product is defined as the sum of the products of the components of each vector as:

F · D=
`(4√(2))*3+(4√(2))*2=20√(2)`

||F||= 8

||D||=
`√(3^2+2^2) =√(13)`

Hence:

Θ=arccos(
`(20√(2) )/(8√(13) )`)

Θ=arccos(0.981)

Θ= 11.3°

c) Work is equal to:

F · D=
`(4√(2))*3+(4√(2))*2=20√(2)`=28.3

Other way of obtainig the work is:

||F||||D||cos(Θ)

where:

||F||= 8

||D||=
`√(3^2+2^2) =√(13)`

Θ=11.3°

So, ||F||||D||cos(Θ)=8×
`√(13)`×cos(11.3°)=28.3