A 6 m board of mass 7 kg is hinged at one end, with a 60 kg block nesting on the

ID: 2302560 • Letter: A

Question

A 6 m board of mass 7 kg is hinged at one end, with a 60 kg block nesting on the board

245 cm from the hings. A force ~F is applied vertically at the other end to lift of the board.

1) Find the magnitude of the force needed to hold the board stationary at at angle of elevation

of 33?. Assume the block is a point particle. The acceleration of gravity is 9.81 m/s2.

Answer in units of N

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

Answer in units of N

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

angle of elevation of the board is 33?.

Answer in units of N

4) Find the force exerted by the hinge if ~F is exerted perpendicular to the board when the

angle of elevation of the board is 33?.

Answer in units of N

A)
let F is the force on other end

net torque about hinge is zero.

7*3**9.81*sin(57) + 60*9.81*2.45*sin(57) - F*6*sin(33) = 0

F = (7*3*9.81*sin(57) + 60*9.81*2.45*sin(57))/(6*sin(33))

= 422.97 N

B) Fnety = 0

Fh - 7*9.81 - 60*9.81 +F = 0

Fh = (7*9.81+60*9.81)- 422.97

= 234.3 N

3.5*9**9.81*sin(55) + 72*9.81*2.22*sin(55) - F*6*sin(90) = 0

F =( 7*3*9.81*sin(57) + 60*9.81*2.45*sin(57)/(6)

= 127.6 N

B) Fnety = 0

Fhy - 7*9.81 - 60*9.81 +F*cos(33) = 0

Fhy = (7*9.81+60*9.81)-127.6*cos(33)

= 550.26 N

Fhx = F*sin(35)

= 127.6*sin(33)

= 69.5 N

Fh = sqrt(Fhx^2+Fhy^2) = 554.63 N

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