The truck bed and provide the deceleration of the crate. Static friction must be backward, to oppose the impending slide along V = v 0 + at 0 = 4.5 m/s + a(3.4 s) which gives a = - 1.32 m/s 2 Using the force diagram for the boy, we can write F The acceleration can be found from the boy's one-dimensional motion:
= m 2g sin 2 Plugging in numbers we get (1.50 kg)sin62° = Two equations together, to eliminate T, we get m 1g sin 1 Since the string doesn't stretch, the tension is the sameīlock, using the force diagrams: Block 1, x-component: m 1g sin 1 - T = 0īlock 2, x-component: T - m 2g sin 2 = 0 since we require the acceleration to be 0. There are two different coodinate systems used, one for each M 2F/(m 1 + m 2) Note that an equal and opposite force acts on m 1. The maximum tension: (85 kg)(9.8 m/s 2) - 700 N = (85 kg)a min which gives a min = 1.56 m/s 2 b) For this constant acceleration, we can write y = y 0 + v 0t + ½at 2 = 0 + 0 + ½(1.6 m/s 2)(4.0 s) 2 = 12.5 m To find the velocity, we use v = v 0 + at = 0 + (1.6 m/s 2)(4.0 s) =Ī) If we choose the two blocks as a system we can writeį = (m 1 + m 2)a which gives a = F/(m 1 + m 2) b) Choosing the block on the right as the system now, then Mg - T = ma From this we see that there is a minimum acceleration that produces As shown in the figure, we take down to be the positiveĭirection (this is the direction of the acceleration). The man will pull down on the rope, and the rope will The tension in the rope is created by the contact between the ropeĪnd the man's hands. Note: Symbols written in Bold are vectors. PHYSICS 101 - Homework 4 - Fall 2000 Physics 101 - Homework # 4 Solutions
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