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

Answer:

a) F= 4$\sqrt{2}$ i + 4$\sqrt{2}$ j

b) Θ=11.3

c) The work done is 20$\sqrt{2}$

Step-by-step explanation:

a) ||F||=8, α=π/4

Fx=||F||·sin(π/4)=8·$\frac{\sqrt{2}}{2}$

Fy=||F||·cos(π/4)=8·$\frac{\sqrt{2}}{2}$

F=Fx i + Fy j = 4$\sqrt{2}$ i + 4$\sqrt{2}$ j

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

cos(Θ)=$\frac{F.D}{||F||||D||}$

where:

D= 3 i + 2 j

F=4$\sqrt{2}$ i + 4$\sqrt{2}$ j

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

F · D= $(4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}$

||F||= 8

||D||= $\sqrt{3^2+2^2} =\sqrt{13}$

Hence:

Θ=arccos($\frac{20\sqrt{2} }{8\sqrt{13} }$)

Θ=arccos(0.981)

Θ= 11.3°

c) Work is equal to:

F · D= $(4\sqrt{2})*3+(4\sqrt{2})*2=20\sqrt{2}$=28.3

Other way of obtainig the work is:

||F||||D||cos(Θ)

where:

||F||= 8

||D||= $\sqrt{3^2+2^2} =\sqrt{13}$

Θ=11.3°

So,  ||F||||D||cos(Θ)=8×$\sqrt{13}$×cos(11.3°)=28.3