Question

Three forces of magnitudes F1=4.0N, F2=6.0N, and F3=8.0N are applied to a block of mass m=2.0kg, initially at rest, at angles shown on the diagram. (Figure 1)In this problem, you will determine the resultant (net) force by combining the three individual force vectors. All angles should be measured counterclockwise from the positive x axis (i.e., all angles are positive).

Part A

Calculate the magnitude of the resultant force \(\vec{F}_{\mathrm{r}}=\vec{F}_{1}+\vec{F}_{2}+\vec{F}_{3}\) acting on the block.

Part B

What angle does \(\vec{F}_{\mathrm{r}}\) make with the positive \(x\) axis?

Part C

What is the magnitude of the block's acceleration vector, \(\vec{a}\) ?

Answer 1

- solution Just have a look at the figure, we start making - componente of Forces Fi Fi and Ğ alorg ad y axes. Note: Iz is already along over anis, no component analysis for this . Now & components along x, Fix = F, cos 25° along y, Fry - Fi Sin 25 and Fa components 325.4 360-325> along , Fax = F Cos 35' along-yaris, Fay = - Eg sin35° and F along x axis, Fax = -F3 and Fzy=0 = 350 .: Fen = Fix + Fax + Fax = Ficos 25' + F Cos 35'- Ez = 0.54 (After substituting F=4N F2=6N F3=8N L Fey = Fly + Fay + Fzy = Fisin25' - Fg sin35' to = -1.75 .: Fe = (Ferfit Eryl Fe = 0.54 î - 1.75) - (1)

.. Part A Feln 10.54)2+(-1.75)2 = 70,2916 +3.0825 Magnitude = 1313541 Fz = 1.831 N upto 2 significant and figures, | Fe = 1.8N Part c Newton's second law, Fe= mã .. magnitude of a = te i 1.8 = 0.9155 & upto a significant figures, (Q1 = 0.15 ml - Part B Now were eq" ("). 8 = 0.54ê - 1075 let I be the angle Fe makes with txaris, Now, Fey is negative and for is positive, Hence Ê lies in fourth quodrant.

Hence using trigonometry, = 21 - tant (IFcyl) - xata 1 l Fal 2= 21 - tant / 1.75 ) 0.54 T= 2(180°) – tant 2= 360' – 72,84 Le 287.152 and upto two significant figures, Id= 290° measured from tue xaxis in counterclockwise sense. my enforts :-) Hope you like solution & my

Answer 2

WRONG factually incorrect my dear sir