Question

For problems 81-83, a ladder stands against a wall (assumed friction is present on the wall). The ladder is 4 m long and makes an angle of 37° with the horizontal at the base. Ewall 81. What if r- 0 about point A? (Sum of clockwise torques about the axis of rotation sum of counterclockwise torques about the same axis of rotation.) A. il x cos 8x4x sin 37)+ mx g x [(4x cos 37)/2]. 423-0, 37) / 2] ng N wall +(Fwall x sin θχ4xcos 37. 37 (Fwall x sin Bx 4 x sin 37) m×g×[(4 × cos 37) / 2]. D. (Fwall x cos θ× 4) +(Fwall x sin θ)-(rn × g) + F,-。 Problems 81-83 82. What if the vector sum of the horizontal force components C. N-Fe What if the vector sum of the vertical force components 83. C. (Fwall x sin θ) + N = m x g http/www.justUSEit.com gm@justu PKT IOB

Answer 1

For problems 81-83, a ladder stands against a wall (assumed friction is present on the wall). The ladder is 4 m long and makes an angle of 37° with the horizontal at the base. Ewall What ifr= 0 about point A? (Sum of clockwise torques about the axis of rotation sum of coupterclockwise torques about the same axis of rotation.) 81. Fwail x cos 8x 4 x sin 37)+ ng mx g x [(4x cos 37)/2]. 10, 37) / 2] N +(Fwall X sin 8x4 x cos 37 (Fwd cos θ×4×cos 37) + 37 (Fwall x sin θ× 4 × sin 37)= mxgx [(4 x cos 37)/2]. Problems 81-83 D. 82. What if the vector sum of the horizontal force components B. C. N-Fe What if the vector sum of the vertical force components F,-07 A. (Fvilx sin θ) + N-Ft. 83. http/www.justUSEit.com gm@justu PKT IOB

the ans I have provided above as we can see

in part a angle $\Theta$ is not given we find it as 53^o as 53+37=90   and sin53 =cos37 and vise versa Now since sin²$\Theta$+cos²$\Theta$= 1 we have the simplifies equation as

F_wall *4=mg*4/2*cos37

also in case of b and c we can see that by applying newton's law we get the solution easily