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yuradex [85]
2 years ago
10

A 0.305 kg book rests at an angle against one side of a bookshelf. The magnitude and direction of the total force exerted on the

book by the left side of the bookshelf are given by: FL= 1.52 NθL=31 degrees What must the magnitude and direction of the total force exerted on the book by the bottom of the bookshelf be in order for the book to remain in this position? Fb= _____Nθb=_____degree
Physics
1 answer:
tankabanditka [31]2 years ago
5 0

Answer

given,

F_L= 1.52\ N

\theta_L= 31^0

mass of book = 0.305 Kg

so, from the diagram attached  below

F_L cos {\theta_L} + F_b sin {\theta_b} = m g

1.52 times cos {31^0} + F_b sin {\theta_b} = 0.305 \times 9.8

F_b sin {\theta_b} = 2.989 -1.303

F_b sin {\theta_b} = 1.686

computing horizontal component

F_b cos {\theta_b} = F_L sin {\theta_L}

cos {\theta_b} = \dfrac{F_L sin {\theta_L}}{F_b}

cos {\theta_b} = \dfrac{1.52 \times sin {31^0}}{1.686}

cos {\theta_b} = 0.464

θ = 62.35°

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To find the constant A and B we have to compare this equation with equation 1.

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Given:-

- The following differential equation for (x) the distance a rain drop has fallen has the form:

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- Where,                v = Speed of the raindrop

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Find:-

(a) Using the proposed solution for v find the acceleration a.

(b) Find the distance the raindrop has fallen in t = 3.00 s.

(c) Given that k = 2.00 g/m, find the mass of the raindrop at t = 3.00 s.

Solution:-

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- Similarly, we know that velocity (v) is the first derivative of displacement (x):

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