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Physics questions （urgent！）
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1. Let T1 and T2 be the tension in the string on boxes A and B respectively. Hence, T1 - 10gsin(37) - Ff = 10a and 10g - T2 = 10a where a is the acceleration of the system, and g is the acceleration due to gravity (taken to be 10 m/s^2) Ff is the friction force But Ff = 0.3 x 10g.cos(37) N = 23.96 N Therefore, T1 = (10a + 10g.sin(37) + 23.96) N = (10a + 84.14) N ---------- (1) and T2 = (10g - 10a) N = (100 - 10a) N ----------- (2) Consider the pulley, use: torque = moment of inertia x angular acceleration (T2 - T1) x (0.05) = 0.1 x (a/0.05) i.e. T2 - T1 = 40a ------------- (3) Using (1) and (2): T2 - T1 = (100 - 10a) - (10a + 84.14) T2 - T1 = 15.86 - 20a Frm (3): 15.86 - 20a = 40a a = 15.86/60 m/s^2 = 0.264 m/s^2 2. Where is the figure ?
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Physics questions （urgent！）
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1）Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. http://zuhairusnizam.uitm.edu.my/lecture%20notes/phy093-chapter%205%20slides.pdf(... 顯示更多 1）Box A, of mass 10.0 kg, rests on a surface inclined at 37° to the horizontal. It is connected by a lightweight cord, which passes over a massless and frictionless pulley, to a second box B, which hangs freely as shown. http://zuhairusnizam.uitm.edu.my/lecture%20notes/phy093-chapter%205%20slides.pdf ( slide 61) If the mass of the pulley cannot be neglected and the pulley has a rotational inertia (around its axle) of 0.1kgm^2,and the coefficient of kinetic friction is 0.30, and mB = 10.0 kg, determine the acceleration of the system. Suppose the pulley rotates frictionless lay around its axle. The radius of the pulley is 5cm. 2) A beam with a length of 8.00 m and a mass of 100. kg is attached by a large bolt to a support at a distance of 3.00 m from one end. The beam makes an angle θ = 30.0° with the horizontal, as shown in the figure. A mass M = 500. kg is attached with a rope to one end of the beam, and a second rope is attached at a right angle to the other end of the beam. (A)Find the tension, T, in the second rope and (ans=2380N) (b)the force exerted on the beam by the bolt. ( ans= 3822N) (C) if the second rope is suddenly cut find the angular acceleration of the beam at the moment when the second rope is broken. (Ans=18.8 rads^-1) I want to know the steps and answers!!! Please help me...最佳解答:
1. Let T1 and T2 be the tension in the string on boxes A and B respectively. Hence, T1 - 10gsin(37) - Ff = 10a and 10g - T2 = 10a where a is the acceleration of the system, and g is the acceleration due to gravity (taken to be 10 m/s^2) Ff is the friction force But Ff = 0.3 x 10g.cos(37) N = 23.96 N Therefore, T1 = (10a + 10g.sin(37) + 23.96) N = (10a + 84.14) N ---------- (1) and T2 = (10g - 10a) N = (100 - 10a) N ----------- (2) Consider the pulley, use: torque = moment of inertia x angular acceleration (T2 - T1) x (0.05) = 0.1 x (a/0.05) i.e. T2 - T1 = 40a ------------- (3) Using (1) and (2): T2 - T1 = (100 - 10a) - (10a + 84.14) T2 - T1 = 15.86 - 20a Frm (3): 15.86 - 20a = 40a a = 15.86/60 m/s^2 = 0.264 m/s^2 2. Where is the figure ?
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