NOTE - the reaction force is not shown on the free body diagram you just drew. Doyou understand why?Perpendicular and ParallelForces.A 500 +/- 1 g aluminum block sits inthe middle of a balance. Strings withsmall mass hangers are attached asshown in the diagram. The masshanger on the right has a total mass of100 +/- 1 g. The hanger on the left has amass of 50 +/- 0.5 g. Assume that thebalance reads 400.1 +/- 0.1g.50gAlum. BlockBalance100gAs before, follow the procedure in thelab manual to obtain a free body diagram and ultimately the second lawequations for the aluminum block. It will be easier if you separate the x and yforces and deal with each direction individually.When you have completed writing the second law equations above, answer thefollowing questions:2. What is the magnitude of the frictional force acting between the aluminum blockand the balance? (You should be able to determine this from your free bodydiagrams.) 3. Calculate the numeric values for the forces given. Using the second law equationyou wrote for the vertical forces, verify if the forces sum to zero within experimentalerror. If not within the expected experimental error, are the values obtained out ofreason considering the apparatus and experimental arrangement?4. You add 50 g of mass to the left mass hanger (for a total mass of 100 g), what is thefrictional force between the aluminum block and the balance?5. Leaving the left mass hanger with a total of 100 g mass, you add 100 g of mass tothe right mass hanger (for a total mass of 200 g). What is the frictional force betweenthe aluminum block and the balance?6. Using a simple sketch, identify the reaction to the normal force acting on thealuminum block. What is its direction and magnitude compared to the normal force?

Perpendicular and Parallel Forces. A 500 +/- 1 g aluminum block sits in the middle of a balance. Strings with small mass hangers are attached as shown in the diagram. The mass hanger on the right has a total mass of 100 +/- 1 g. The hanger on the left has a mass of 50 +/-0.5 g. Assume that the balance reads 400.1 +/-0.1g. 50g Alum. Block Balance 100g As before, follow the procedure in the lab manual to obtain a free body diagram and ultimately the second law equations for the aluminum block. It will be easier if you separate the x and y forces and deal with each direction individually.

Perpendicular and Parallel Forces. A 500 +/- 1 g aluminum block sits in the middle of a balance. Strings with small mass hangers are attached as shown in the diagram. The mass hanger on the right has a total mass of 100 +/- 1 g. The hanger on the left has a mass of 50 +/-0.5 g. Assume that the balance reads 400.1 +/-0.1g. 50g Alum. Block Balance 100g As before, follow the procedure in the lab manual to obtain a free body diagram and ultimately the second law equations for the aluminum block. It will be easier if you separate the x and y forces and deal with each direction individually.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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