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2 answers:
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Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
Answer:
F= 600 N
Explanation:
Given that
Initial velocity ,u= 0 m/s
Final velocity ,v= 30 m/s
mass ,m = 0.5 kg
time ,t= 0.025 s
The change in the linear momentum is given as
ΔP= m (v - u)
ΔP= 0.5 ( 30 - 0 ) kg.m/s
ΔP= 15 kg.m/s
We know that from second law of Newtons
Now by putting the values
F= 600 N