Hydraulic press is used to lift a load of 500 kg when the ratio of the diameter of pump plunger and press plunger is 1:5. The effort applied on piston of plunger . If the mechanical advantage of handle of pump plunger is 4,the effort applied on handle of pump plunge is
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I have given the diagram for ur question u can check that:
Given:
F1=mg[force will be in upward direction and weight of the body downward]
So,,,
F1=500 into 10[m=500kg and g=10m/s^2]
F1=5000 N
F2=?
A1=πr^2
=πd^2/2^2
A2=πr^2
=πd^2/2^2
d1=5m
d2=1m
According to the given data we can apply the formula:
F1/A1=F2/A2
5000/π(d1/2)2=F2/π(d2/2)2
5000/5^2=F2/1^2
5000/25=F2
F2=200 N
Given:
F1=mg[force will be in upward direction and weight of the body downward]
So,,,
F1=500 into 10[m=500kg and g=10m/s^2]
F1=5000 N
F2=?
A1=πr^2
=πd^2/2^2
A2=πr^2
=πd^2/2^2
d1=5m
d2=1m
According to the given data we can apply the formula:
F1/A1=F2/A2
5000/π(d1/2)2=F2/π(d2/2)2
5000/5^2=F2/1^2
5000/25=F2
F2=200 N
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