A square steel plate is 10.0 cm on a side and 0.500 cm thick. (a) Find the shear strain that results if a force of magnitude 9.0×10^5 N is applied to each of the four sides, parallel to the side. (b) Find the displacement x (in centimeters).

Video transcript

Welcome back everybody. We are looking at an aluminum box here and we are told that a tangential force is applied on both ends of this aluminum box. Now this tangential force is equal to three times 10 to the fifth. Newtons, we are tasked with finding two different things here. We are tasked with finding what the shear strain of the box is and we are tasked with finding what the displacement of the boxes. And we are also told just as I note that the dimension of the box is six by six by six centimeters. So let's start out here with the shear strain. Well, we know that the sheer module is which I'm just going to denote by S is equal to the tangential force over the area of the cross section of the box, divided by the displacement over the height of one of the sides of the boxes. Now the displacement over the height is actually equal to our shear strain. So our sheer modulates, then is equal to force over area over our shear strain, which rearranging this equation. We then get that our shear strain is equal to our tangential force over our area of a cross section of the boss box. Times our sheer modulates which we know all these values. So let's go ahead and plug those values in. Our shear strain is equal to our tangential force of three times 10 to the fifth, divided by area of a cross section. Cross section of a cube is just a square. So we're just gonna do six centimeters or 60.6 m squared times the sheer module list for aluminum, which when you look it up is just 2.5 times 10 to the 10th pascal's. Which when you plug this into your calculator, we get that our shear strain is 3.3 times 10 to the negative third. Great. Now moving on to part two here, we need to find the displacement. Well, we established earlier that our shear strain is equal to displacement over height, meaning that our displacement is just equal to our shear strain times the height of one of the sides of the box is plugging in values. Here we get that. This is 3.3 times 10 to the negative third times a height of one of the sides, which is six centimeters or 0.6 m. Which when you plug this into your calculator, you get 0.198 micro meters. So now we have found both the shear strain and the displacement which corresponds to uh Sorry, answer choice. C Thank you all so much for watching. Hope this video helped. We will see you all in the next one