Answer:
a) τ₁ = 660 N m, b) τ’= 686 N m, c) F = 623.6 N
Explanation:
a) For this exercise let's use the concepts of torque and rotational balance.
For this we set a reference system at the base and assuming that the counterclockwise rotations are positive
where the force F = 600 N, the distance to the axis is x = 1.1 m, the mass of the system m = 70g and the weight is placed at the point of the center of gravity x_{cm} = -1.0 m
The torque at the front is
τ₁ = F x
τ₁ = 600 1.1
τ₁ = 660 N m
b) let's write the rotational equilibrium condition
∑ τ = 0
τ'- W x_{cm} = 0
τ ’= mg x_{cm}
τ’= 70 9.8 1.0
τ’= 686 N m
c) the greatest force Matt can apply
τ’= F x
F = τ’/ x
F = 686 / 1.1
F = 623.6 N
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Answer:
7167.27N/m
Explanation:
To get the force constant k, we will use the formula;
F = kx where
F is the force acting on the object down the plane
k is the spring constant
x is the extension
The force acting down the ramp us F = Wsintheta. The formula will become;
Wsintheta = kx
mgsintheta = kx
K = mgsintheta/x
K = (800)(9.8)sin47°/0.8
K= 7840sin47°/0.8
K = 5733.81/0.8
K = 7167.27N/m
Hence the spring constant of the spring is 7167.27N/m
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Explanation: