Physics 206 Example Problems Newton's Laws of Motion

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$MATH 219 - Differential Equations Quiz 3 Name Solutions Date ...$

That is I can conclude that the tensions in the two sides of the ropes are equal. Applying the second law in the y-direction gives, Fnet, y = m a y T1 sinθ + T2 sinθ − F = 0 2 T sinθ − F = 0 T = F 2 sinθ = 400 2 sin( 3◦ N = 3820 N ) The size of the force that the rope pull on the car with is the tension in the rope which has just been found to be 3820 N. B) To just move the car the acceleration has to be barely more that zero. So the net force is essentially zero still. The analysis of part A) still holds so, T = F 2 sinθ = 600 2 sin( 4◦ N = 4300 N ) Problem 1 1 . What size downward force F must be applied to lift the cart of weight 2000 N using the 4pulley apparatus shown below? F m Ceiling Solution . The tension in the rope going over the pulleys is constant throughout. Apply Newton’ s second law to each of the hanging pulleys. The upward force on each of these pulleys is 2 F. Since these pulleys do not accelerate ( and if the pulleys are nearly massless. . . ) the downward force on these pulleys must have size 2 F. This means that the tensions in each of the strings attached to the mass is 2 F. Therefore the net upwards force on the mass is 4 F. This force supports the weight of the mass so, 4 F = W ⇒ F = W 2000 N = = 500 N 4 4 To lift the mass a force of 1 /4 the weight of the block must be applied. The pulley arrangement is essentially a force multiplier. Problem 1 2. A mass m hangs at the end of a massless string attached to the roof of a truck. The motion is such that the string makes a constant angle θ with the vertical. What must be the acceleration of the truck for this situation to be possible. ( This forms the basis for a simple means to measure the acceleration of an object. ) Solution: The forces that act on the mass are the tension in the string and the force due to the gravitational pull of the earth. With up as +y and the direction of the acceleration as +x, these forces can be written as, Frop e = T sinθ xˆ + T cosθyˆ FG = − m gE yˆ Applying Newton’ s second law in the y-direction gives, Fnet, y = m a y = 0 T cosθ − m gE = 0 m gE T = cosθ 8
Applying the second law in the x-direction gives, T sinθ = m ax T sinθ ax = m = ( m gE/cosθ) sinθ = gE tanθ m So, measuring the angle θ allows for a determination of the acceleration of the vehicle. 9
Page 1 and 2: Problem 1 . Physics 206 Example Pro
Page 3 and 4: 7. Plugging in the numbers: Fpy = (
Page 5 and 6: Solution : There is a range of mass
Page 7: I have assumed directions for the r

Physics 206 Example Problems Newton's Laws of Motion

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