Statics Worksheet 1 Answers 1. A worker that has a mass of 100 kg is standing on scaffolding whose supports are 5.0 m apart. He is standing on a uniform beam of mass 50.0 kg. (a) Find the force on each support. 5.0 m Wt2 F1 F2 Wt1 Wt1 = m1 g = ( 100 kg )( 9.8 m/s2 ) = 980 N Wt2 = m2 g = ( 50 kg )( 9.8 m/s2 ) = 490 N

5.0 m 490 N F1 F2 980 N Forces up = Forces down F1 + F2 = 980 + 490 (1) F1 + F2 = 1470

5.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 Clockwise torques = Counter-clockwise torques τCW = τCCW Need a pivot point about which to apply the torques; can choose any point we wish; choose the pivot point to be at the left end of the beam

5.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 Clockwise torques: from the beam and the worker Counter-clockwise torques: from F2 Note: F1 does not produce a torque, since its torque arm is zero

5.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 τCW = τCCW τ = r F ( 2.5 m )( 980 N ) + ( 2.5 m )( 490 N ) = ( 5 m ) F2 3675 m N = ( 5 m ) F2 F2 = 735 N

5.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 (1) F1 + F2 = 1470 F2 = 735 N F1 + 735 = 1470 F1 = 735 N

(b) If the man moves to a point 2 (b) If the man moves to a point 2.0 m from the left end, what is the force on each support? 5.0 m 2.0 m 490 N F1 F2 980 N Forces up = Forces down F1 + F2 = 980 + 490 (1) F1 + F2 = 1470

5.0 m 2.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 τCW = τCCW τ = r F ( 2 m )( 980 N ) + ( 2.5 m )( 490 N ) = ( 5 m ) F2 3185 m N = ( 5 m ) F2 F2 = 637 N

5.0 m 2.0 m 490 N F1 F2 980 N (1) F1 + F2 = 1470 (1) F1 + F2 = 1470 F2 = 637 N F1 + 637 = 1470 F1 = 833 N

what is the force on each support? (c) If a load of bricks of mass 400 kg is placed 2.0 m from the right end, what is the force on each support? Wt. = ( 40 kg )( 9.8 m/s2 ) = 392 N 5.0 m 2.0 m 2.0 m F1 392 N 490 N F2 980 N Forces up = Forces down F1 + F2 = 980 + 490 + 392 (1) F1 + F2 = 1862

5.0 m 2.0 m 2.0 m F1 392 N 490 N F2 980 N (1) F1 + F2 = 1862 τCW = τCCW τ = r F (2 m)(980 N) + (2.5 m)(490 N) + (3.0 m)(392 N) = (5 m) F2 4361 m N = (5 m) F2 F2 = 872 N

5.0 m 2.0 m 2.0 m F1 392 N 490 N F2 980 N (1) F1 + F2 = 1862 (1) F1 + F2 = 1862 F2 = 872 N F1 + 872 = 1862 F1 = 990 N

and again 3.0 m from that end. At the end hangs a sign that 2. A cantilever of length 8.0 m and mass 1000 kg is supported at one end and again 3.0 m from that end. At the end hangs a sign that weighs 300 N. Find the force on each support. Wt. = ( 1000 kg )( 9.8 m/s2 ) = 9800 N 8.0 m 3.0 m EAT F2 F1 9800 N 300 N Forces up = Forces down F1 + F2 = 9800 + 300 (1) F1 + F2 = 10 100

8.0 m 3.0 m EAT F2 F1 9800 N 300 N Clockwise torques = Counter-clockwise torques Clockwise torques: from the beam and the sign Counter-clockwise torques: from F2

8.0 m 3.0 m EAT F2 F1 9800 N 300 N τCW = τCCW τ = r F ( 4 m )( 9800 N ) + ( 8.0 m )( 300 N ) = ( 3 m ) F2 41 600 m N = ( 3 m ) F2 F2 = 13 867 N

8.0 m 3.0 m EAT F2 F1 9800 N 300 N (1) F1 + F2 = 10 100 F2 = 13 867 N F1 + 13 867 = 10 100 F1 = - 3767 N F1 must pull down on the beam in order for equilibrium to be established

3. Find the forces in the diagram below. The system is in equilibrium. 2 m 1 m 1 m 1 m F1 40 N F2 50 N 75 N 85 N Forces up = Forces down F1 + F2 = 50 + 75 + 40 + 85 (1) F1 + F2 = 250

(1 m)(50 N) + (3 m)(75 N) + (4 m )(40 N) + (5 m)(85 N) = (6 m) F2 2 m 1 m 1 m 1 m F1 40 N F2 50 N 75 N 85 N τCW = τCCW τ = r F (1 m)(50 N) + (3 m)(75 N) + (4 m )(40 N) + (5 m)(85 N) = (6 m) F2 860 m N = ( 6 m ) F2 F2 = 143 N

1 m 2 m 1 m 1 m 1 m F1 40 N F2 50 N 75 N 85 N (1) F1 + F2 = 250 F2 = 143 N F1 + 143 = 250 F1 = 107 N