A wood block is sliding up a wood ramp. if the ramp is very steep, the block will reverse direction at its highest point and sli
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Answer:
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Explanation:
Given data,
The mass of the bar AB, m = 6.4 kg
The angular velocity of the bar, θ˙ = 2.7 rad/s
The angle of the bar at A, θ = 24°
Let the length of the bar be, L = l
The angular moment at point A is,
∑ Mₐ = Iα
Where, Mₐ - the moment about A
α - angular acceleration
I - moment of inertia of the rod AB
Let G be the center of gravity of the bar AB
The position vector at A with respect to the origin at G is,
The acceleration at the center of the bar
Since the point A is fixed, acceleration is 0
The acceleration with respect to the coordinate axes is,
Comparing the coefficients of i
Comparing coefficients of j
Net force on x direction
substituting the values
=1.5(14.58L+11.96)
Similarly net force on y direction
= 3.2(2.97L - 157.03) + 62.72
Where L is the length of the bar AB
Therefore the net force,
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Substituting the value of L gives the force at pin A