Question

T = Mgx.mx - m.magd = 0) = m = Mxmax – (9500 kg)(3.4 m) - 426 9)(3.4 m) – 4200 kg d (7.7 kg) 5. - 2.0 m .- (a) Let m = 0. Calculate the net torque about the left end of the diving board, with counterclockwise torques positive. Since the board is in equilibrium, the net torque is zero. 1.0 m mţ Mi t = F (1.0 m) - Mg (4.0 m)= 0 → F = 4 Mg = 4( 52 kg)(9.80 m/s") = 2038 N = 2.0x10'N, up ------- 4.0 m ------ Use Newton's second law in the vertical direction to find FA F = Fs - Mg – F = 0 → F = Fs - Mg = 4 Mg - Mg = 3Mg = 3 ( 52 kg)(9.80 m/s) = 1529 N ~ 1500 N, down (b) Repeat the basic process, but with m= 28 kg. The weight of the board will add more clockwise torque. Et = F (1.0 m) – mg (2.0 m) – Mg (4.0 m) = 0) ► Fn = 4 Mg +2mg = [4(52 kg) + 2(28 kg)](9.80 m/s?) = 2587 N = 2600 N, up F = F - Mg - mg - Fi → F = Fg - Mg - mg = 4Mg +2mg - Mg - mg = 3Mg + mg = [3(52 kg) + 28 kg](9.80 m/s) = 1803 N = 1800 N, down 6. Write Newton's second law for the junction, in both the x and y directions F = F,-F, cos 45º = 0 From this, we see that F > F Thus set F. = 1660 N. mg

#5 where Fa=Fb-Mg=+4Mg-Mg.
where the positive 4 I believe they moved from left to right between the = sign. but then where the -Mg is coming from?

Answer 1

The force F downwards The force Fz upwards And weight Mg downwards At equilibrium, the net vertical force is zero EF, = 0 F,-F,- Mg = 0 -Fi = Mg-F F = Fg - Mg = 4Mg - Mg = 3 Mg In this he calculated F and F3 =4Mg Here Mg and F, have negative sign because they are in negative y-direction