Two carts with masses of 4.32 kg and 2.66 kg move toward each other on a frictionless track with speeds of 4.86 m/s and 3.93 m/s
1 answer:
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Answer:
a) 29062.125 N·m
b) 0 N·m
c)
d) T = 11186.02 N
Explanation:
We are given
Beam mass = 1975 kg
Beam length = 3 m
Cable angle = 60° above horizontal
a) We have the formula for torque given as follows;
Torque about the pin = Force × Perpendicular distance of force from pin
Where the force = Force due to gravity or weight, we have
Weight = Mass × Acceleration due to gravity = 1975 × 9.81 = 19374.75 N
Point of action of force = Midpoint for a uniform beam = length/2
∴ Point of action of force = 3/2 = 1.5 m
Torque due to gravity = 19374.75 N × 1.5 m = 29062.125 N·m
b) Torque about the pinned end due to the contact forces between the pin and the beam is given by the following relation;
Since the distance from pin to the contact forces between the pin and the beam is 0, the torque which is force multiplied by perpendicular distance is also 0 N·m
c) To find the expression for the tension force, T we find the sum of the moment forces about the pin as follows
Sum of moments about p is given as follows
∑M = 0 gives;
T·sin(θ) × L= M×L/2×g
Therefore torque due to tension is given by the following expression
d) Plugging in the values in the torque due to tension equation, we have;
Therefore, we make the tension force, T the subject of the formula hence