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1. A small block of mass 25 kg is on a rough slope inclined at α degree to the horizontal. The block is held in equilibrium by a force of magnitude P N applied paralled to the slope and up the slope.When P = 259, the block is abput to slip up the slope. When P = 35, the block is about to slip down the... 顯示更多 1. A small block of mass 25 kg is on a rough slope inclined at α degree to the horizontal. The block is held in equilibrium by a force of magnitude P N applied paralled to the slope and up the slope. When P = 259, the block is abput to slip up the slope. When P = 35, the block is about to slip down the slope. In each case, the magnitude of the frictional force acting on the block is F N. i.) Calculate the vlue of α and show that F = 112. ii.) Calcaulate the coefficient of friction between the block and the slope. 2. The diagram shows a mass of 50 kg on a slope which makes an angle of 30 degree with the horizontal. The coefficient of friction between the mass and the slope is 0.25. You may assume that the mass does not tip up. The diagram :: http://i121.photobucket.com/albums/o209/jenny_0531/1.jpg Find the magnitude of the force T if i.) the mass is about to move down the slope ii.) the mass is about to move up the slope iii.) the mass is accelerating at 5 ms^-2 up the slope.
最佳解答:
1.(i) When the block is about to slide up the slope, frictional force is pointing downward along the slope 259 = 25g.sin(alpha) + Ff ---------- (1) where Ff is the frictional force When the block is about to slip down the slope, frictional force is pointing upward along the slope, 35 + Ff = 25g.sin(alpha) i.e. 35 = 25g.sin(alpha) – Ff ---------- (2) (1)+(2): 294 = 50g.sin(alpha) solving for (alpha) gives (alpha) = 36 degrees (1)-(2): 224 = 2Ff i.e. Ff = 112 N (ii) Coefficient of friction = 112/25g.cos(36) = 0.55 2. Consider direction normal to the slope, 50g.cos(30) = T.sin(20) + R, where R is the normal reaction hence, R = 50g.cos(30) – T.sin(20) Friction Ff = 0.25R = 0.25.( 50g.cos(30) – T.sin(20)) -------- (3) (i) Friction is pointing upward along the slope, T.cos(20) + Ff = 50g.sin(30) Substitute Ff from (3) and solve for T (ii) Friction is pointing downward along the slope, T.cos(20) = 50g.sin(30) + Ff Substitute Ff from (3) and solve for T (iii) Friction is again pointing downward along the slope. Use [net force] = mass x acceleration T.cos(20) – [50g.sin(30) + Ff] = 50 x 5 Substitute Ff from (3) and solve for T
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Friction - Mechanics 2 急>發問:
1. A small block of mass 25 kg is on a rough slope inclined at α degree to the horizontal. The block is held in equilibrium by a force of magnitude P N applied paralled to the slope and up the slope.When P = 259, the block is abput to slip up the slope. When P = 35, the block is about to slip down the... 顯示更多 1. A small block of mass 25 kg is on a rough slope inclined at α degree to the horizontal. The block is held in equilibrium by a force of magnitude P N applied paralled to the slope and up the slope. When P = 259, the block is abput to slip up the slope. When P = 35, the block is about to slip down the slope. In each case, the magnitude of the frictional force acting on the block is F N. i.) Calculate the vlue of α and show that F = 112. ii.) Calcaulate the coefficient of friction between the block and the slope. 2. The diagram shows a mass of 50 kg on a slope which makes an angle of 30 degree with the horizontal. The coefficient of friction between the mass and the slope is 0.25. You may assume that the mass does not tip up. The diagram :: http://i121.photobucket.com/albums/o209/jenny_0531/1.jpg Find the magnitude of the force T if i.) the mass is about to move down the slope ii.) the mass is about to move up the slope iii.) the mass is accelerating at 5 ms^-2 up the slope.
最佳解答:
1.(i) When the block is about to slide up the slope, frictional force is pointing downward along the slope 259 = 25g.sin(alpha) + Ff ---------- (1) where Ff is the frictional force When the block is about to slip down the slope, frictional force is pointing upward along the slope, 35 + Ff = 25g.sin(alpha) i.e. 35 = 25g.sin(alpha) – Ff ---------- (2) (1)+(2): 294 = 50g.sin(alpha) solving for (alpha) gives (alpha) = 36 degrees (1)-(2): 224 = 2Ff i.e. Ff = 112 N (ii) Coefficient of friction = 112/25g.cos(36) = 0.55 2. Consider direction normal to the slope, 50g.cos(30) = T.sin(20) + R, where R is the normal reaction hence, R = 50g.cos(30) – T.sin(20) Friction Ff = 0.25R = 0.25.( 50g.cos(30) – T.sin(20)) -------- (3) (i) Friction is pointing upward along the slope, T.cos(20) + Ff = 50g.sin(30) Substitute Ff from (3) and solve for T (ii) Friction is pointing downward along the slope, T.cos(20) = 50g.sin(30) + Ff Substitute Ff from (3) and solve for T (iii) Friction is again pointing downward along the slope. Use [net force] = mass x acceleration T.cos(20) – [50g.sin(30) + Ff] = 50 x 5 Substitute Ff from (3) and solve for T
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