A stone that has a mass of m = 6.5 kg rests on a horizontal plane. The coefficie

Question

A stone that has a mass of m = 6.5 kg rests on a horizontal plane. The coefficient of static friction, s, is 0.35. A horizontal force, F, is applied to the stone, and it is just enough to get the stone to begin moving.

(690) Problem 9: A stone that has a mass of m = 6.5 kg rests on a horizontal plane The coefficient of static friction, us is 0.35. A horizontal force, F, is applied to the stone, and it is just enough to get the stone to begin moving ©theexpertta.com 20% Part (a) Choose the correct Free Body Diagram from the choices below given that F is the static friction force, Fr is the normal force and Fg is the weight 20% Part (b) Write an expression for the sum of the forces in the x-direction using the variables from the above Free Body Diagram Grade Summa Deductions Potential 2% 9896 7 8 9 Submissions Attempts r (19% per attempt) detailed view emaining: 18 F. 1% 1% Submit I give up! Hints: 3 for a 0% deduction. Hints remaining: 0 Feedback: 1% deduction per feedback. Which forces act in the x-direction? Are the forces both oriented along x? Are they both oriented along-x? You do not need the static coefficient of friction for this part. Write the Second Newton's law for the x-direction 20% Part (c) Given the coordinate system specified in the problem statement, write an expression for the sum of the forces in the y- direction. - 20% Part (d) Write an expression to show the relationship between the maximum friction force, Ff and the normal force, FN - 20% Part (e) Calculate the magnitude of F, in Newtons, if Fris at its maximum.

Explanation / Answer

(B) Fx = F - Ff

(C) Fy = FN - Fg

(d) Ff = us FN

(e) block is at rest.

Fx = F - Ff = 0

F = Ff = us FN

and Fy = FN - Fg = 0

FN = 6.5 x 9.8

F = 0.35 x 6.5 x 9.8

F = 22.3 N

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