A cat walks along a uniform plank that is 4.20 m long and has a mass of 6.00 kg

Question

A cat walks along a uniform plank that is 4.20 m long and has a mass of 6.00 kg . The plank is supported by two sawhorses, one 0.440 m from the left end of the board and the other 1.50 m from its right end. If the cat has a mass of 3.3 kg, how close to the right end of the two-by-four can it walk before the board begins to tip? I understand that to get the answer you ultimately use d=M/m(distance from the sawhorse to the cat), but I do not understand how to get the distance from the sawhorse to the cat. A cat walks along a uniform plank that is 4.20 m long and has a mass of 6.00 kg . The plank is supported by two sawhorses, one 0.440 m from the left end of the board and the other 1.50 m from its right end. If the cat has a mass of 3.3 kg, how close to the right end of the two-by-four can it walk before the board begins to tip? I understand that to get the answer you ultimately use d=M/m(distance from the sawhorse to the cat), but I do not understand how to get the distance from the sawhorse to the cat.

Explanation / Answer

Let the needed distance be L. (from the right saw-horse to the cat). The torque about the right saw-horse is zero. so clock wise. F*2,26+3,3*g*L (2,26 is the distance of the two saw-horses) counter clock-wise 6*g*0,6. (0,6 is the distance from the center of gravity of the plank to the right saw-horse) so we have. F*2,26+3,3*g*L=6*g*0,6. so 3,3*g*L=3,6g-2,26*F. we can see that if F=0 then L is maximum. so L=3,6g/3,3g=1,09(m).

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