25x = 200*30
25x= 6000
x= 240
The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.
Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)
N = the normal reaction, which is equal to
N = mg cos(12°)
R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)
F = the component of the weight acting down the incline. It is
F = mg sin(12°)
Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
Therefore, the static coefficient of friction is
μ = tan(12) = 0.213
Answer: μ = 0.21 (nearest tenth)
I expect that they will <em>add</em>, and their effect at every location will be the <em>sum</em> of their individual effects at that location.
For example:
If they're acting at the same point and in opposite directions, the effect will be the same as a single force at that point, with strength equal to their difference, and in the direction corresponding to whichever one is stronger.
Answer:
we have to find out the critical resolved shear stress. As it it given in the question
Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.
a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.
cos(62.4°) = 0.46
cos(72.0°) = 0.31
cos(81.1°) = 0.15
Thus, the slip direction is at the angle of 62.4° along the tensile axis.
b) now the critical resolved shear stress can be find out by the following equation.
τ = σ ( cosФ cosλ)
now by putting values,
= (1.95MPa)[ cos(28.1) cos(62.4)] = 0.80 MPa (114 Psi) 7.23