An object in equilibrium has three forces exerted on it. a 42 n force acts at 90° from the x-axis and a 46 n force acts at 60° f
rom the x-axis. what are the magnitude and direction of the third force? n ° (counterclockwise from the +x direction)
1 answer:
Since the object is in equilibrium, then the sum of the forces acting on it is zero. You must do the sum of forces for the component "y" and for the component "x". From there you clear the components of the third force. Then you must find the direction and magnitude.
The result is a force of 85.01N at 74.30 degrees from the x axis. Attached solution.
The result is a force of 85.01N at 74.30 degrees from the x axis. Attached solution.
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Answer:
Explanation:
= normal force acting on the coin
Normal force in the upward direction balances the weight of the coin, hence
= frequency of rotation
Angular velocity of turntable is hence given as
= distance from the axis of rotation
= minimum coefficient of static friction
static frictional force is given as
The static frictional force provides the necessary centripetal force , hence
Centripetal force = Static frictional force
Answer:
v_{1fy} = - 0.4549 m / s
Explanation:
This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved
initial. Before the crash
p₀ = m v₁₀
final. After the crash
= m
+ m v_{2f}
Recall that velocities are a vector so it has x and y components
p₀ = p_{f}
we write this equation for each axis
X axis
m v₁₀ = m v_{1fx} + m v_{2fx}
Y Axis
0 = -m v_{1fy} + m v_{2fy}
the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components
sin 23.3 = v_{2fy} / v_{2f}
cos 23.3 = v_{2fx} / v_{2f}
v_{2fy} = v_{2f} sin 23.3
v_{2fx} = v_{2f} cos 23.3
we substitute in the momentum conservation equation
m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3
0 = - m v_{1f} sin θ + m v_{2f} sin 23.3
1.83 = v_{1f} cos θ + 1.15 cos 23.3
0 = - v_{1f} sin θ + 1.15 sin 23.3
1.83 = v_{1f} cos θ + 1.0562
0 = - v_{1f} sin θ + 0.4549
v_{1f} sin θ = 0.4549
v_{1f} cos θ = -0.7738
we divide these two equations
tan θ = - 0.5878
θ = tan-1 (-0.5878)
θ = -30.45º
we substitute in one of the two and find the final velocity of the incident ball
v_{1f} cos (-30.45) = - 0.7738
v_{1f} = -0.7738 / cos 30.45
v_{1f} = -0.8976 m / s
the component and this speed is
v_{1fy} = v1f sin θ
v_{1fy} = 0.8976 sin (30.45)
v_{1fy} = - 0.4549 m / s