Thank you for the help. 113.8 N is wrong and I can\'t figure this one out. (8%)

Question

Thank you for the help. 113.8 N is wrong and I can't figure this one out.

(8%) Problem 7: Two blocks are pressed together as shown. The blocks have masses mi 545 kg and m2-10.1 kg. The static coefficient of friction between the two blocks is 0.47·The tabletop is frictionless. A rightward force, F, is exerted on Block 1 . 2 frictionless tabletop × Calculate the minimum force (in Newtons) required to keep Block 1 from slipping along Block 2. Grade Summary F= 113.81 2% 98% Potential sinO | cosO | tano (1 ) | 7| 8| 9|HOME cotan asinO acos0 atanO acotan0sinh0 coshO tanhO cotanh0+ Submissions Attempts remaining: 5 % per attempt) detailed view 2% 0 END °Degrees Radians VO BACKSPACE EL CLEAR Submit I give up! Hints: 0-for a-deduction. Hints remaining:- Feedback: 0% deduction per feedback.

Explanation / Answer

Let the force be F
Hence,

Frictional Force = 0.47*F

Weight of m1 = 5.45

To balance m1
0.47*F = 5.45*9.8
F = 113.638 N

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