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...
Physics questions （urgent！）
2013-05-22 1:17 am
回答 (1)
2013-05-22 9:18 pm
✔ 最佳答案
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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