Question

A book with a mass of 2.0 kg is held in equilibrium on a board with a slope of 60.0° by a horizontal foce. What is the normal force exerted on the book?

Answer 1

The normal force exerted on the book is 9.8067 N, which is calculated using the mass of the book, the acceleration due to gravity, and the cosine of the slope angle (60°).

Step-by-step explanation:

To determine the normal force exerted on the book, we need to consider the forces at play when a book is held on an inclined plane with a slope of 60.0°. The weight of the book (mass times the acceleration due to gravity) acts vertically downward. This weight can be broken down into two components: one perpendicular to the slope (which contributes to the normal force) and one parallel to the slope. The parallel component is counteracted by the horizontal force that keeps the book in equilibrium, and hence does not affect the normal force. The component of the weight perpendicular to the slope is what contributes to the normal force. Thus, with a mass of 2.0 kg and using the value of the acceleration due to gravity (9.8067 m/s²), the normal force can be calculated as the product of the mass, the acceleration due to gravity, and the cosine of the slope angle, which is 60.0° in this case.

As the book is in equilibrium, the normal force is equal to the weight component perpendicular to the slope. The force due to gravity is Fg = m × g, where m is mass and g is acceleration due to gravity. Therefore:

• Calculate the force due to gravity: Fg = 2.0 kg × 9.8067 m/s² = 19.6134 N
• Determine the perpendicular component: Fg × cos(60°) = 19.6134 N × 0.5 = 9.8067 N
• The normal force is thus 9.8067 N.

Answer 2

When three forces are in equilibrium at a point, their ratios are the same as those of the corresponding lengths in a triangle formed from their three directions.

I'll work in kg-equivalent-masses first, converting to N (Newtons) only at the end.

The kg-equivalent forces are those of the weight due to a mass of 2.0 kg acting downwards (short "perpendicular side" of triangle), of
2 sqrt(3) kg horizontally (long perp. side) and of 4.0 kg along the hypoteneuse, normal to the board.
(The ratios of these sides, 1sqrt(3)2 are the ratios for the 30, 60, 90 deg. triangle.)

So the normal force exerted by the book is the force due to the weight of a 4.0 kg mass, that is
4.0 x 9.8 N =
And the answer is
39 N.
Have a nice day