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A force is applied to a block sliding along a surface (Figure 2). The magnitude of the force is 15 N, and the horizontal component of the force is 4.5 N. At what angle (in degrees) above the horizontal is the force directed?

A force is applied to a block sliding along a surface Figure 2 The magnitude of the force is 15 N and the horizontal component of the force is 45 N At what angl class=

Respuesta :

If I have done my math correctly, your answer is 67.5

Explanation:

It is given that,

The magnitude of force applied force on the block is, F = 15 N

The horizontal component of the force, [tex]F_x=4.5\ N[/tex]

We need to find the angle (in degrees) above the horizontal is the force directed. The horizontal component of the force is given by :

[tex]F_x=F\ cos\theta[/tex]

[tex]4.5=15\ cos\theta[/tex]    

[tex]\theta=cos^{-1}(\dfrac{4.5}{15})[/tex]

[tex]\theta=72.54^{\circ}[/tex]

So, the angle above the horizontal force and the surface is 72.54 degrees. Hence, this is the required solution.

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