Answered

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the crate is drawn accross the floor by a winch that retracts the cable at a constant rate of 0.2m/s. the crates mass is 120kg, the coefficient of kinetic friction between the crate and the floor is 0.24 at instant shown, what is the tension in the cable?

Sagot :

**For convenience imagine the cable retracts the crate to the RIGHT 
constant rate of 0.2m/s, a constant rate of motion means no acceleration,
   thus a = 0
-the crate is in dynamic equilibrium-

Using the formula of
[tex]net Force = ma[/tex]
[tex]net Force = m(0)[/tex]
[tex]net Force = 0N[/tex]
Then since the net force, in this case is the Force to the right MINUS the force of Friction thus
[tex]F - f = 0[/tex]
[tex]F = f[/tex]

Because the crate is on a flat surface, the normal force = weight
Weight = m g
Weight = 120 * 9.81
W = 1772.2 N

First calculate the force of friction using the formula
[tex]f = Fn (friction coefficient)[/tex]
[tex]f = Fn (0.24)[/tex]
[tex]f = 1177.2(0.24)[/tex]
[tex]f=282.528N[/tex]

now use this is in the equation before
[tex]F=f[/tex]
[tex]F = 282.528N[/tex]