A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensile axis. Three possible slip directions make angles of 62.4 72.0 , and 81.1 with the same tensile axis.(a) Which of these three slip directions is most favored?(b) If plastic deformation begins at a tensile stress of 1.95 MPa (280 psi), determine the critical resolved shear stress for aluminum.

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

the force required to move the cart is divided by three or 33.33 n.

i am pretty sure it is c because when people are closer together, they cannot move as freely. picture being in a mostly full elevator, for instance. the people are right next to you, and you can barely move.