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圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437382.jpg For Q8,I don't know how to calculate the torque about O for option A,C and D.How to calculate them?For Q10(a),below shows the answer圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437393.jpg The moment of... 顯示更多 圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437382.jpg For Q8,I don't know how to calculate the torque about O for option A,C and D.How to calculate them? For Q10(a),below shows the answer 圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437393.jpg The moment of weight of the object acts on M instead of A.Why takes A as the pivot? For Q11,I don't know how to do
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8. Torque = force x perpendicular distance between the force and pivot. Option A: Torque = 2 x 3 + (-2 x 3) Nm = 0 Nm (The perpendicular distance between the force 2N and the pivot is 3 units) Option C: Torque = (2x1 + 2x4) Nm = 10 Nm Option D: Torque = (2x4 + (-2x3)) Nm = 2 Nm 10. Just imagine if the mass is too heavy, the stand would topple about the point A. That is to say, A is the centre of rotation of the whole system. 11. The motion of the block follows the equation, F - Ff = ma where Ff is the max frictional force m is the mass of the block a is the acceleration hence, a = (F - Ff)/m Using equation of motion: s = ut + (1/2)at^2 with u = 0 m/s, a = (F-Ff)/m, t = 2 s hence, s = (1/2).[(F-Ff)/m].(2^2) i.e. s = 2(F-Ff)/m or s = (2/m).F - 2Ff/m This is an equation of a straight line with slope (2/m) and x-intercept Ff, which is just the graph shown in option B.
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圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437382.jpg For Q8,I don't know how to calculate the torque about O for option A,C and D.How to calculate them?For Q10(a),below shows the answer圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437393.jpg The moment of... 顯示更多 圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437382.jpg For Q8,I don't know how to calculate the torque about O for option A,C and D.How to calculate them? For Q10(a),below shows the answer 圖片參考：http://imgcld.yimg.com/8/n/HA04937267/o/701202150048313873437393.jpg The moment of weight of the object acts on M instead of A.Why takes A as the pivot? For Q11,I don't know how to do
最佳解答:
8. Torque = force x perpendicular distance between the force and pivot. Option A: Torque = 2 x 3 + (-2 x 3) Nm = 0 Nm (The perpendicular distance between the force 2N and the pivot is 3 units) Option C: Torque = (2x1 + 2x4) Nm = 10 Nm Option D: Torque = (2x4 + (-2x3)) Nm = 2 Nm 10. Just imagine if the mass is too heavy, the stand would topple about the point A. That is to say, A is the centre of rotation of the whole system. 11. The motion of the block follows the equation, F - Ff = ma where Ff is the max frictional force m is the mass of the block a is the acceleration hence, a = (F - Ff)/m Using equation of motion: s = ut + (1/2)at^2 with u = 0 m/s, a = (F-Ff)/m, t = 2 s hence, s = (1/2).[(F-Ff)/m].(2^2) i.e. s = 2(F-Ff)/m or s = (2/m).F - 2Ff/m This is an equation of a straight line with slope (2/m) and x-intercept Ff, which is just the graph shown in option B.
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