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
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.
To determine the coefficient of static friction, we can use the formula:
Coefficient of static friction (μ) = Force of friction (Ff) / Normal force (Fn)
1. Given that the block of wood weighs 10 Newton (W = 10 N) and the spring balance records a horizontal force of 5 Newton (F = 5 N) when the block just begins to move, we can assume that the force of friction just equals the force applied. Therefore Ff = 5 N.
2. The normal force (Fn) is the force exerted by the table on the block in the upward direction. Since the block is on a horizontal table and is not accelerating vertically, the normal force equals the weight of the block. Hence Fn = W = 10 N.
Now we can calculate the coefficient of static friction:
μ = Ff / Fn = 5 N / 10 N = 0.5
Therefore, the coefficient of static friction is 0.5.
To find the friction force when the spring balance records a horizontal force of 8 Newton (F = 8 N), we can use the formula:
Friction force (Ff) = μ * Normal force (Fn)
We already know μ = 0.5, and Fn = 10 N.
Ff = 0.5 * 10 N = 5 N
Therefore, the friction force when a horizontal force of 8 Newton is recorded by the spring balance is 5 Newton.