(10%) Problem 8: A horizontal force, F, = 95 N, and a force, F2 = 19.1 N acting

Question

(10%) Problem 8: A horizontal force, F, = 95 N, and a force, F2 = 19.1 N acting at an angle of to the horizontal, are applied to a block of mass m-2.8 kg. The coefficient of kinetic friction between the block and the surface is ,-02. The block is moving to the right. Randomized Variables F1 =95 N F2=19.1 N m=2.8 kg Otheexpertta.com 50% Part (a) Solve numerically for the magnitude of the normal force. FN in Newtons, that acts on the block if@m 30". Grade Summary Potential 100% Submissions Ghi per attempt sin() cos() tan() | | ( : 7 8 9 cotanO asin0 acos0 atan0 acotan sinho cosh tan tanh0t detailed vievw 0 grees Radians WO Igive up deduction per feedback. Submit Hint Feedback: ' 50% Part (b) Solve numerically for the magnitude of acceleration of the block, a in m/s2, if = 30°.

Explanation / Answer

a)Total normal reaction = weight + due to F2

= mg + F2 sin(theta)

=2.8*9.81 + 19.1*0.5

=37.018 N

b)Net frictional force = u*N

=0.2*37.018

=7.4036 N

so, let the acceleration be a.so,

balancing the horizontal forces,

F1 - f - F2 cos(30) = ma

or 2.8*a = 95 - 7.4036 - 2.94

or a=30.23 m/s^2

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