(29%) Problem 5: A horizontal force, F 75 N, and a force, F 12.6 N acting at an
ID: 1597528 • Letter: #
Question
(29%) Problem 5: A horizontal force, F 75 N, and a force, F 12.6 N acting at an angle of 0 to the horizontal, are applied to a block of mass m 3.3 kg. The coefficient of kinetic friction between the block and the surface is uk 0.2.The block is moving to the right Randomized variables Fu J 75 N F2 12.6 N m 3.3 kg Otheexpertta.com 50% Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if 0 30° Grade Summary FNEH Deductions 0% Potential 100% tan() C 7 8 9 HOME Submissions cos0 sin Attempts remaining: 7 per attempt) cotan0 asin acos0 4 5 6 detailed view 1 2 3 atan() acotan0 sinh() cosh() tanh0 cotanh() Degrees Radians BACKSPACE CLEAR Submit Feedback: 0% deduction per feedback. A 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if 0 30Explanation / Answer
a)
normal force, N=mg+F2*sin(theta)
N=(3.3*9.8+12.6*sin(30)
N=38.64 N
b)
m*a=F1-F2*cos(theta)-uk*N
3.3*a=75-12.6*cos(30)-0.2*38.64
===> a=17.08 m/sec^2
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