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A contestant in a winter sporting event pushes a block of ice of mass m across a frozen lake as shown in the figure. The coefficient of static friction between the block and ice is μs, and the coefficient of kinetic friction is μk. θ is the angle the force makes with the x-axis.
In this part, we are going to set-up Newton's second Law equations for the cases
(1) when the ice block just starts moving, and
(2) when it is accelerating to the right with an acceleration a.
All answers are symbolic. ALL ANSWERS ARE CASE-SENSITIVE.
Subpart 1: Newton's Second Law along the y-axis
(i) Write Newton's Second Law along the y-axis by adding all forces in the y-direction taking into account their signs (forces pointing upwards are positive and downward are negative) in terms of the normal force N, weight mg, F and θ.
In both scenarios, there is no acceleration along the y-direction, therefore, ay=0.
∑Fy=
= m ay = 0 (1)
(ii) Using (1) to solve for N.
N=
(2)
Think: In (2) is N greater than the weight, less than the weight or equal to the weight?
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