An 8 m board of mass 5 kg is hinged at one end, with a 56 kg blocknesting on the

Question

An 8 m board of mass 5 kg is hinged at one end, with a 56 kg blocknesting on the board 262 cm from the hings. A force F is appliedvertically at the other end to lift of the board. F = [ mL + 2x/2L ]g *costheta Find the force exerted by the hinge if ~F is exerted perpendicularto the board when the angle of elevation of the board is 28?.Answer in units of N. need the last part for thisquestion Find the magnitude of the force needed to hold the board stationaryat at angle of elevation of 28?. Assume the block is apoint particle. The acceleration of gravity is 9.81 m/s^2 . Answerin units of N. I got this to be 204.232 N, which wascorrect Find the force exerted by the hinge at thisangle. Answer in units of N. I got this to be 393.568, which was correct Find the magnitude ofthe force similar F if similar F is exerted perpendicular to the board when theangle of elevation of the board is 28?. Answer in units ofN. I tried this equation and got itwrong,

Explanation / Answer

The third and fourth part of the question should be exactly thesame as the first and second, except without a cos(28) attached toF. In other words - balancing torques, we get - 56 * 9.81 * cos(28) * 2.62    +    4* 5 * 9.81cos(28)    = F * 8 F = (56 * 9.81 * cos(28) * 2.62   +    4 * 5 * 9.81cos(28)) / 8 F = 180.5 N Another way to think about it: Take your original force, 204.2, andmultiply it by cos(28). This gives you the perpendicular componentof the original force. Since torque only takes into accountperpendicular components, this is your new answer. I am not nearly as confident with this part as theprevious part, but here we go: Force of hinge: Sum of up forces (with x axis along board) = Fhu + 180.5 =56*9.81cos(28) + 5 * 9.81cos(28) Fhu = 347.865 Sum of sideways forces = Fhs =56*9.81sin(28) + 5 * 9.81sin(28) Fhs = 280.936 Pythagorean theorem, and you get 447.14 N

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