A lead brick with the dimensions shown in the figure below (6.0 cm left to right, 5.0 cm front to back and 2.0 cm top to bottom) rests on a rough, solid surface. A force of F = 2300 N is applied as indicated, with θ = 29°.
(a) Find the change in height of the brick.
(b) Find the amount of shear deformation.
F=2300 N Fsinθ = 2300xsin29O = 1115 N
Fcosθ = 2300xcos29O
=2012 N l = 0.06m, b = 0.05m, h = 0.02m
Force F can be resolved into horizontal and vertical components, Fcosθ and Fsinθ respectively.
The vertical component causes compression of the lead brick and the horizontal ...
The expert finds the force and change in height and amount of shear deformation. A step by step solution provided.