Question

Four forces act on an object, given by A = 40 N East, B = 50 N North, C = 70 N West, and D = 90 N South. What is the magnitude of the net force on the object? Explain your answer

Answer 1

Answer:

Solution is given in attachment

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Answer 2

Concept Introduction:

A body can change its condition of rest or motion by the application of force, which is an external agent. It has a size and a movement. The force is applied in both a direction and at a point, which is referred to as the force's direction and application, respectively.

Given:

We have been given:

Four forces act on an object, given by

$A = 40 N east$

$B = 50 N North$

$C = 70 N West$

$D = 90 N South$

To Find:

We have to Find:

The magnitude of the net force on the object

Solution:

According to the problem,

The net force has components,

$F_{x} = A_{x} + B_{x} + C_{x} + D_{x}$ = $40+0-70+0 = - 30 N$

And, $F_{y} = A_{y} + B_{y} + C_{y} + D_{y}$ = $0+50+0-90= -40 N$

The magnitude of the net force is,

$F^{2} = F^{2} _{x} + F_{y}^{2}$ = $(-30)^{2} + (-40)^{2}$

Hence, $F= 50 N$

Final Answer:

The net force on the object is $50 N$

SPJ2