(3%) Problem 24: A horizontal force, F,-75 N, and a force, F2 = 17.3 N acting at
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(3%) Problem 24: A horizontal force, F,-75 N, and a force, F2 = 17.3 N acting at an angle of to the horizontal, are applied to a block of mass m1 = 3.7 kg. The coefficient of kinetic friction between the block and the surface is 0.2. The block is moving to the right Randomized Variables F1=75 N F2= 17.3 N rn = 3.7 kg ©theexpertta.com 50% Part (a) Solve numerically for the magnitude of the normal force, FN in Newtons, that acts on the block if = 30°. Grade Summary Deductions Potential FN- 0% 100% tan() | | ( | cosO cotan0 asin acos0 atanO acotan)sinh(0 cosh0 tanh) cotanh0 Degrees ORadians Submissions Attempts remaining: 2 (2% per attempt) detailed view sinO 0 END BACKSPACE DEL CLEAR Submit Hint I give up! Hints: 0% deduction per hint. Hi ints remaining: Feedback: 0% deduction per feedback. 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s, if = 30°.Explanation / Answer
Given,
F1 = 75 N ; F2 = 17.3 N ; m = 3.7 kg ;
theta = 30 deg ; uk = 0.2
a)The normal force acting on the block will be:
N = F2 sin(theta) + mg
N = 17.3 sin30 + 3.7 x 9.8 = 44.91 N
hence, N = 44.91 N
b)The frictional force on the block is:
Ff = uk N = 0.2 x 44.91 = 8.98 N
The net force is;
Fnet = F1 - (Ff + F2 cos(theta))
Fnet = 75 - (8.98 + 17.3 cos30)) = 51.04 N
Fnet = ma = 51.04 =>a = 51.04/m
a = 51.04/3.7 = 13.79 m/s^2
Henc, a = 13.79 m/s^2
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