A block of mass 20kg is pushed against a vertical surface at an angle of 30 degrees as shown. The coefficient of friction between the surface and the block is 0.2. What is the minimum magitude of P to hold the block still?
I got 271.5 when I tried but the answer is 202.2
image link - http://img80.imageshack.us/img80/9630/mjlmwu5.jpg
2007-02-28 11:31:58 · 3 answers · asked by bollocks 2 in Science & Mathematics ➔ Physics
The force F should be divided into its component parallel to the surface and perpendicular to the surface (the normal force).
The parallel force, Fy = F cos 30°.
The perpendicular force, Fx = Fn = F sin 30° (normal force)
Call the coefficienf ot friction "f"
The frictional force, (Fn)(f), must equal the force on the block due to gravity minus the upward force Fy.
The force down is 20kg x 9.8 m/s² = 196 N
So the force up must be:
Fy + (Fn)(f) = 196 N
F cos 30° + (0.2)F sin 30° = 196
0.866F + (0.2)(0.5)F = 196
0.966F = 196
F = 196/0.966 = 202.9 N
2007-02-28 11:56:57 · answer #1 · answered by CheeseHead 2 · 0⤊ 0⤋
How did you get that?
The force supporting the block is P·cos(30°) + µ·P·sin(30°). You set that equal to m·g and solve for P.
2007-02-28 11:55:55 · answer #2 · answered by injanier 7 · 0⤊ 0⤋
you be responsive to that the rigidity of friction = coeffiient of friction * widely used rigidity. what's the rigidity of friction? nicely, if a horizontal rigidity of 56N keeps the crate shifting with consistent speed (0 acceleration). If the rigidity shifting the crate became into larger than the rigidity of friction, it may enhance up, if it became into much less, it may decelerate, so the rigidity of friction = the rigidity to maintain it shifting. What with regard to the traditional rigidity? Any merchandise sitting on a flat, point floor, the traditional rigidity = m*g. So, you be responsive to the traditional rigidity and the rigidity of friction, in simple terms remedy for the coefficient of kinetic friction.
2016-10-02 03:24:18 · answer #3 · answered by catharine 3 · 0⤊ 0⤋