A 7 m board of mass 9 kg is hinged at one end, with a 72 kg block resting on the

ID: 2302591 • Letter: A

Question

A 7 m board of mass 9 kg is hinged at one end, with a 72 kg block resting on the board 222 cm from the hinges.

A force ~F is applied vertically at the other end to lift the board.

A) Find the magnitude of the force needed to hold the board stationary at at angle of eleva-

tion of 35. Assume the block is a point particle. The acceleration of gravity is 9.81 m/s2 .

Answer in units of N

B) Find the force exerted by the hinge at this angle.

Answer in units of N

C) Find the magnitude of the force ~F if ~F is exerted perpendicular to the board when the

angle of elevation of the board is 35. Answer in units of N

D) Find the force exerted by the hinge if ~Fis exerted perpendicular to the board when the

angle of elevation of the board is 35. Answer in units of N

A)
let F is the force on other end

net torque about hinge is zero.

3.5*9**9.81*sin(55) + 72*9.81*2.22*sin(55) - F*7*sin(35) = 0

F = (3.5*9**9.81*sin(55) + 72*9.812.22*sin(55))/(7*sin(35))

= 382.96 N

B) Fnety = 0

Fh - 9*9.81 - 72*9.81 +F = 0

Fh = (9*9.81+72*9.81)-381.96

= 412.65 N

3.5*9**9.81*sin(55) + 72*9.81*2.22*sin(55) - F*7*sin(90) = 0

F = (3.5*9**9.81*sin(55) + 72*9.812.22*sin(55))/(7)

= 219.66 N

B) Fnety = 0

Fhy - 9*9.81 - 72*9.81 +F*cos(35) = 0

Fhy = (9*9.81+72*9.81)-219.66*cos(35)

= 614.68 N

Fhx = F*sin(35)

= 219.66*sin(35)

= 126 N

Fh = sqrt(Fhx^2+Fhy^2) = 652.74 N

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.