Question #274335

Joey and his friends are pushing a piece of driftwood which they will be using in their landscaping project. They pushed the wood by applying the following forces:

Joey: 5 N, north

Friend 1: 2.2 N, west

Friend 2: 3.86 N, 55° north of west

What is the net force applied by Jerry and his friends on the wood?


1
Expert's answer
2021-12-02T10:04:49-0500

Let's find xx- and yy-components of the net force:

Fnet,x=5 N×cos90+2.2 N×cos180+3.86 N×cos(18055)=4.41 N,F_{net,x}=5\ N\times cos90^{\circ}+2.2\ N\times cos180^{\circ}+3.86\ N\times cos(180^{\circ}-55^{\circ})=-4.41\ N,Fnet,y=5 N×sin90+2.2 N×sin180+3.86 N×sin(18055)=8.16 N.F_{net,y}=5\ N\times sin90^{\circ}+2.2\ N\times sin180^{\circ}+3.86\ N\times sin(180^{\circ}-55^{\circ})=8.16\ N.

We can find the magnitude of the net force from the Pythagorean theorem:


Fnet=Fnet,x2+Fnet,y2,F_{net}=\sqrt{F_{net,x}^2+F_{net,y}^2},Fnet=(4.41 N)2+(8.16 N)2=9.27 N.F_{net}=\sqrt{(-4.41\ N)^2+(8.16\ N)^2}=9.27\ N.

We can find the direction of the net force from the geometry:


θ=tan1(FyFx),\theta=tan^{-1}(\dfrac{F_y}{F_x}),θ=tan1(8.16 N4.41 N)=62 N of W.\theta=tan^{-1}(\dfrac{8.16\ N}{-4.41\ N})=62^{\circ}\ N\ of\ W.

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