Answer:
Velocity of truck will be 20.287 m /sec
Explanation:
We have given mass of the truck m = 4000 kg
Radius of the turn r = 70 m
Coefficient of friction
Centripetal force is given
And frictional force is equal to
For body to be move these two forces must be equal
So
The tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
<h3>Tension in the cable</h3>
Apply the principle of moment and calculate the tension in the cable;
Clockwise torque = TL sinθ
Anticlockwise torque = ¹/₂WL
TL sinθ = ¹/₂WL
T sinθ = ¹/₂W
T = (W)/(2 sinθ)
T = (29 x 9.8)/(2 x sin57)
T = 169.43 N
<h3>Vertical component of the force</h3>
T + F = W
F = W - T
F = (9.8 x 29) - 169.43
F = 114.77 N
Thus, the tension in the cable is 169.43 N and the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.
Learn more about tension here: brainly.com/question/24994188
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Answer:
I just toughed it out and talked with friends
Explanation: