A 0.400 kg bead slides on a straight frictionless wire and moves with a velocity of 3.50 cm/s to the right, as shown below. The
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Answer:
x = 0.176 m
Explanation:
For this exercise we will take the condition of rotational equilibrium, where the reference system is located on the far left and the wire on the far right. We assume that counterclockwise turns are positive.
Let's use trigonometry to decompose the tension
sin 60 = / T
T_{y} = T sin 60
cos 60 = Tₓ / T
Tₓ = T cos 60
we apply the equation
∑ τ = 0
-W L / 2 - w x + T_{y} L = 0
the length of the bar is L = 6m
-Mg 6/2 - m g x + T sin 60 6 = 0
x = (6 T sin 60 - 3 M g) / mg
let's calculate
let's use the maximum tension that resists the cable T = 900 N
x = (6 900 sin 60 - 3 200 9.8) / (700 9.8)
x = (4676 - 5880) / 6860
x = - 0.176 m
Therefore the block can be up to 0.176m to keep the system in balance.
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its <span>r=radius,B=angle in rad
velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and
Now we can go to proportion
SOLVING FOR A :
or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
Do not forget about </span> w = angular speed in rad/s and
Now we can go to proportion
SOLVING FOR A :
</span>I'm sure it helps!