Question

An object experiences two forces acting on the same surface (i.e. they are additive): a force of 5.0N acting at 60.∘ to the horizontal and a force of 8.0N acting 45∘ to the horizontal.

What is the magnitude of the resultant force on the object? Give your answer to the nearest newton, without units.

Answer 1

Answer:

Helo chuki kaha thum hhhhhelo brta

Answer 2

The magnitude of the resultant force on the object is 12.71 N.

Given,

Number of forces acting on the surface=2

Force, F1=5.0N acting at 60° to the horizontal

Force, F2=8.0N acting 45° to the horizontal.

To find,

the magnitude of the resultant force on the object.

Solution:

• As they act on the same surfaces, it means that they can be added.
• We need to take components of the forces here.

For force, F1 which is equal to 5.0 N is acting at an angle of 60° to the horizontal:

Let component of force on x-axis and y-axis be Fx and Fy respectively.

Fx=Fcosθ

Fy=Fsinθ.

where,

F-force acting

θ-angle with the horizontal.

Fx=5xcos60°

$Fx=5X\frac{1}{2}\\Fx=\frac{5}{2} N.$

Fy=Fsin60°

$Fy=5X\frac{\sqrt{3} }{2} \\Fy=\frac{5\sqrt{3} }{2} N.$

For force, F2 which is equal to 8.0 N is acting at an angle of 45° to the horizontal:

Fx=5xcos45°

$Fx=8X\frac{1}{\sqrt{2} }\\Fx=\frac{8}{\sqrt{2} } N.$

Fy=Fsin45°

$Fy=8X\frac{1}{\sqrt{2} } \\Fy=\frac{8}{\sqrt{2} } N.$

$Fy=8X\frac{1}{\sqrt{2} } \\Fy=\frac{8}{\sqrt{2} } N.$

The components on x-axis and y-axis will be added respectively.

$(Fx) net=\frac{5}{2} +\frac{8}{\sqrt{2} } \\(Fx) net=\frac{5}{2} +\frac{8\sqrt{2} }{2 }\\(Fx) net=\frac{5+8\sqrt{2} }{2} \\(Fx) net=\frac{5+8X1.4 }{2}\\(Fx) net=\frac{5+11.2}{2}\\(Fx) net=\frac{16.2}{2}\\(Fx)net=8.1N.$

$(Fy) net=\frac{5\sqrt{3} }{2} +\frac{8}{\sqrt{2} } \\(Fy) net=\frac{5\sqrt{3} }{2} +\frac{8\sqrt{2} }{2 }\\(Fy) net=\frac{5\sqrt{3}+8\sqrt{2}}{2}\\(Fy) net=\frac{5X1.7+8X1.4}{2}\\(Fy) net=\frac{8.5+11.2}{2}\\(Fy) net=\frac{19.7}{2}\\(Fy) net=9.85 N\\(Fy)net=9.8 N.$

As the angle between the (Fx)net and (Fy)net vectors is 90°, their resultant will be:

$Fnet=\sqrt{(Fx)net^{2}+(Fy)net ^{2} } \\Fnet=\sqrt{8.1^{2}+9.8^{2} } \\Fnet=\sqrt{65.61+96.04}\\Fnet=\sqrt{161.65}\\Fnet=12.714 N\\Fnet=12.71 N.$

Hence, the magnitude is 12.71 N.

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