Bar (1) has a length of l1 = 2.50 m, and bar (2) has a length of l2 = 0.65 m. initially, there is a gap of d = 3.5 mm between the rigid plate at b and bar (2). after application of the loads p to the rigid plate at b, the rigid plate moved to the right, stretching bar (1) and compressing bar (2). the normal strain in bar (1) was measured as 2,740 μm/m after the loads p were applied. determine the normal strain produced in bar (2).

trig 15.3 cos

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm