Recent content by mhayer7

sorry so if i did it like this since i have 500 and sin 35 using that i can get the perpindicular force which is also the hypoteneuse . so sin35 = 500/z . z = 871.7 then plugging that into your equation we get 871times 1.5 times sin35 = .25sin35 F solving for F we get 5230.3 dividing...

hey drizzle, so using your formula, Fmax[1.5]sin35o=F[0.25]sin35o we wud end up with 500 which is the force applied by the man 500[1.5]sin35o/.25[sin35]= 3000 N 3000N / 9.81 = 305.8 kg's i hope that's right

What is the maximum mass of rocks one can lift using a wheelbarrow PLEASE HELP! Homework Statement A man is using a wheelbarrow to transport rocks across the garden. The man can generate a maximal vertical force of 500N to lift the wheelbarrow. If the centre of mass of the rocks is .25m...