A 4.94-kg block is placed on top of a 11.9-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.505. What is the maximum horizontal force that can be applied before the 4.94-kg block begins to slip relative to the 11.9-kg block, if the force is applied to the more massive block? can someone plz help me...
I've tried doing it by N= 4.94*9.8
=0.505*(4.94)*(9.8) and that is not the right answer can some1 tell me a diff way to do it?
I've tried doing it by N= 4.94*9.8
=0.505*(4.94)*(9.8) and that is not the right answer can some1 tell me a diff way to do it?
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Total of masses = (4.94 + 11.9) = 16.84kg.
(4.94 x 9.8) x 0.505 = friction force of 24.45N.
That is the maximum accelerating force that can be transmitted to the 4.94 kg. block, before it will shift.
Acceleration = (f/m), = 24.54/4.94, = 4.9676m/sec^2. max.
Now find the force required to accelerate both masses together at this rate.
Force = (ma), = 16.84 x 4.9676, = 83.6544N., when applied to the more massive block.
You can round it as you see fit.
(4.94 x 9.8) x 0.505 = friction force of 24.45N.
That is the maximum accelerating force that can be transmitted to the 4.94 kg. block, before it will shift.
Acceleration = (f/m), = 24.54/4.94, = 4.9676m/sec^2. max.
Now find the force required to accelerate both masses together at this rate.
Force = (ma), = 16.84 x 4.9676, = 83.6544N., when applied to the more massive block.
You can round it as you see fit.