Problem 13: Two blocks are tied together with a massless string that does not stretch and connected via a frictionless and massless pulley. Mass one, m1, sits stationary on table top. Denote the tension in the string as T, and the force of static friction as Fs. m2 Part (a) Please use the interactive area below to draw the Free Body Diagram for the mass mj. Add Force y Reset All v Fn 45 X Fs Fg,1 Fn a V m

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A: Visualize the Problem:

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Q: d magnitude of the other forces o

A: Using force conservation law∑Fx=0    and ∑Fy=0From ∑Fx=0 we get 300−Bx=0Bx=300lb(<−)

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Q: Find the Resultant of the following Forces by Component Method. F, = 12 N, West F2 = 24 N, 30° North…

A: Given forces are, F1=12 N, West F2=24 N, 30o North of West F3=15 N, 75o South of West F4=28 N, 50o…

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Q: Given A=173 cm, and B=128 cm, find the coefficient of static friction in the table below. Table 25.…

A: Given: height A =173 cmBase B=128 cm

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A: Approach to solving the question: Detailed explanation:Examples: Key references:

Q: Find the components force in joint * .E,D shown in the figure

A: Given:-

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A: Given: α=30oγ=20oθ=60oμB=0.22μc=0.11μD=0.05p=72.34 NWA=103.45 Ng=9.81 m/s2β=θ+γβ=60+20=80o

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Q: HonzoTital pia ne WiTiCIT Can J t ti Exercise 3-A frictional force of 540N keeps a box stationary on…

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A: Yes, We must always use both the x and the y for every analysis.

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Newton’s Third Law of Motion

The horse pulls the cart and the cart pulls the horse in response as per Newton third law. So the big question is, “Don’t the two forces cancel each other? How does the cart accelerate?”