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Chapter 12: Q87P (page 353)

A particle is acted on by forces given, in newtons, by F1→=8.40i^−5.70j^and F2→=16.0i^+4.10j^ . (a) What are the xcomponentF3→ and (b) ycomponent of the force Fthat balances the sum of these forces? (c) What angle does F3→have relative to the+xaxis ?

Short Answer

Expert verified

a) the x component of force F3, Fx=−24.4.

b) y component of force F3, Fy=1.60.

c) The angle of force F3 with the +x axis,θ=−3.75° .

Step by step solution

01

Understanding the given information

F1→=8.40−5.70

F2→=16.0+4.10

02

 Step 2: Concept and formula used in the given question

Using the equation of equilibrium of force, you can find the components of force F3 and its angle. The formulas are given below.

F→net=F1→+F2→θ=(FyFx)

03

(a) Calculation for thex component  F3→

The net force acting on the particle is given by,

F→net=F1→+F2→

So, Fnet→=24.4i^−1.60j^

Hence, the forceF3→which balances the forceFnet→can be written as,

F3→=−24.4i^+1.60j^

From the equation ofF3→,we get the x component as,

Fx=−24.4 N

04

(b) Calculation for they component of the force F that balances the sum of these forces

From the equation ofF3→you get the y component as,

Fy=1.60 N

05

(c) Calculation for theangle does  F3→ have relative to the  +x-axis

The angle of the forceF3→can be calculated using the equation,

θ=(FyFx)=(1.60 N−24.2 ​N)=−3.75°

The negative sign implies that the angle is measured in the clockwise direction with respect to the positive x-axis.

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