Three blocks, each having a mass of 1.01 kg, are connected by rigid rods of negl
ID: 1956529 • Letter: T
Question
Three blocks, each having a mass of 1.01 kg, are connected by rigid rods of negligible mass and are supported by a frictionless surface. Forces F1 and F2, of magnitude 5.50 N and 11.6 N respectively, are applied to the ends of the blocks as shown in the figure below. Find the forces acting on block B (the +x direction is to the right).The magnitude of the force applied by the rod on the left is?
The magnitude of the force applied by the rod on the right is?
HOW DO I DO THIS PROBLEM? SHOW STEPS PLEASE. I DONT UNDERSTAND
Explanation / Answer
Hello, The easiest way to understand this type of problem is to imagine that you can sit on each of the blocks and look forward and backward. As you look forward, you can only see what is in front of you. As you look backward, you can only see what is behind you. But first we need to cover all blocks with a towel and look at the entire assembly! The total mass = 3 * 1.01 = 3.03 kg The total force = +5.50 + +11.6 = +17.1 N to the right Total force = total mass * acceleration of the total mass 17.1 = 3.03 * a acceleration of the total mass = 16.85 ÷ 3.63= 5.644 Since the rods are rigid, the each block is accelerating at +(5.644) m/s^2 (to the right) NOW it is time to imagine that you can sit on each of the blocks and look forward and backward. If you are sitting on block A, you see F2, +11.6 N, behind you and the rigid rod on the left in front of you. acceleration = +(5.644) m/s^2 mass = 1.01 kg Net Force = 1.01 * +(5.644) = +5.7 N Net force = F2 + Force of left rod +5.7 = +11.6 + F left rod F left rod = 5.7 – 11.6 = -5.9 N This means the left end of the left rod is exerting force of -5.9 N (to the left). Since the rod is rigid, so the right end of the left rod is exerting a +5.9 N force on the left side of block B. NOW you move to block B. If you are sitting on block B, you see a +6.483 N behind you and the rigid rod on the right in front of you. acceleration = +(5.644) m/s^2 mass = 1.01 kg Net Force = 1.01 * +(5.644) = +5.7 N Net force = +6.483 + Force of right rod +5.7 = +5.9 + Force of right rod F right rod =- 0.02 N This means the left end of the right rod is exerting force of -0.02 N on block B (to the left) The magnitude of the force applied by the rod on the left = +.9 N The magnitude of the force applied by the rod on the right is = -0.02 N http://www.webassign.net/colfunphys1/4-p-046.gif This negative sign means the rods felt compressed, so the rods responded by exerting equal forces in the opposite direction on the blocks to which they are attached. Best Regards!
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