Asked by pearl
A block of wood weighing 10newton is placed on the horizontal table,it is then pulled by means of a spring balance attached to one of its ends,the block just begins to move when the spring balance records the horizontal force 5newton.what is the coefficient of static motion and what is the friction force when a horizontal force of 8newton is recorded by the spring balance
Answers
Answered by
Henry
Fb = 10N @ 0deg.
Fh = 5N. = Hor Force.
Fv = 10cos(0) = = 10N. = Force perpendicular to the table.
1. Fn = Fh - Fs = 0,
5 - Fs = 0,
Fs = 5N. = Force of static friction.
u = Fs/Fv = 5/10 = 0.50 = Coefficient of static friction.
Fh = 5N. = Hor Force.
Fv = 10cos(0) = = 10N. = Force perpendicular to the table.
1. Fn = Fh - Fs = 0,
5 - Fs = 0,
Fs = 5N. = Force of static friction.
u = Fs/Fv = 5/10 = 0.50 = Coefficient of static friction.
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