# A string is attached to a 4.5kg block and it is pulling at the block at an angle of 30 degrees above the horizontal. Determine the magnitude and direction of the normal force on the box.

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### 2 Answers

You need to attach a coordinate system to the box that is located on the inclined plane. The x axis will be parallel to the inclined surface and the y axis is perpendicular to the inclined surface.

The weight vector has the origin into the origin of this system of coordinates and it is perpendicular on the ground. You need to decompose the weight vector into its parallel and perpendicular components. The parallel component `G_x` is located along x axis and the perpendicular component `G_y` is located along y axis.

Since the problem does not provides the information concerning the friction force, the forces that act on the box are the tension in spring, the weight force and the normal force.

The normal force acts opposite to the perpendicular component of weight force, such that:

`F_n = G_y = m*g*cos 30^o`

m represents the mass of the box

g represents the gravitational acceleration

`F_n = 4.5*9.8*sqrt3/2 => F_n = 38.19N`

**Hence, evaluating the magnitude of the normal force yields `F_n = 38.19N` and evaluating the direction of the normal force yields that it acts opposite to the perpendicular component of weight force.**

See images :)

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