naveed

Thanx MacSwell, It means its not possible to find force at right sensor by taking left sensor force as a reference? Naveed I calculated the force at right sensor by taking left sensors force as a reference, I calculated more than 9N , I calculating using trignometry (sin(theta) = P/H and cos(theta)=B/H). My expectation was less than 7.01N as right sensor is far away from point A. I dont know the reason that why it is wrong, may be distance shall be involved. Can you comment plz. Regards, Naveed

Its made up of glass. Yes it is rigid but if we push it from any point it will push force sensors. The left and right sides of screen are between aluminum. At the top corners of screen I have mounted force sensors. So if you will push it from back it will push force sensors and we will see the readings. You can refer the attached picture. In picture, red circles are the sensors which are between aluminum and glass screen. I am not sure about the weight but I know the size (Width = 46.5" , Height = 35") Thx dear, Naveed

Practicle/Real Setup: I have a glass screen of length 35" and width 46.5". I have mounted one force sensor at the left top of glass screen and another sensor on the right top of the glass screen. Now, if I press my glass screen from any point on the screen I shall get some force at the left and some force at the right sensor output accordingly. I have hard and tought time to solve this problem. Consider the figure attached, I pressed at point A with (x,y) = (3,32). At the left sensor I am getting 7.015 Newtons. How much force I shall expect/get at right sensor?? Please refer the below info as well. Point A with left sensor: Angle = 45 degrees Distance from sensor = 4.24" Point B with right sensor: Angle = 86.06 degrees Distance from sensor = 43.6" You can calculate angle and distance with your own calculation if I am wrong. Thanks alot in advance, Best Regards, Naveed