(3%) Problem 21: A horizontal force, F 65 N, and a force, F 12.3 N acting at an
ID: 1397970 • Letter: #
Question
(3%) Problem 21: A horizontal force, F 65 N, and a force, F 12.3 N acting at an angle of 0 to the horizontal, are applied to a block of mass m 4Akg.The coefficient of kinetic friction between the block and the surface is Auk 0.2.The block is moving to the right Randomized Variables Fi -65N F2 12.3 N m 4A kg A 50% Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if 0-300 Grade Summary Deductions Potential 100% Submissions 5 Attempts remaining: 20 acos() cotan() asin (09% per attemp) atano acotano sinh0 1 3 detailed view cosh00 tanh() cotanh() vo p Degrees o Radians I give up! Hints: 2 for a deduction. Hints remaining: Feedback 0% deduction per feedback. -Draw a Free Body Diagram of the box. What is the y-component of F2? other than F2 and the normal what other force that acts along y? Submission History Hints Feedback 0% i A 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if 0 30Explanation / Answer
given,
F1 = 65 N
F2 = 12.3 N
theta = 30 degree
mass of the block = 4.4 kg
coefficient of kinetic friction = 0.2
magnitude of the normal force = weight of the block + vertical component of force F2
magnitude of the normal force = mass of block * g + F2 * sin(30)
magnitude of the normal force = 4.4 * 9.8 + 12.3 * sin(30)
magnitude of the normal force = 49.27 N
net horizontal force = F1 - F2 * cos(30)
net horizontal force = 65 - 12.3 * cos(30)
net horizontal force = 54.348 N
force = mass * acceleration
54.348 = 4.4 * acceleration
acceleration = 12.352 m/s^2
acceleration of the block = 12.352 m/s^2
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.