A small box is held in place against a rough wall by someone pushing on it with a force directed upward at 28 degres above the horizontal. the coefficients of static and kinetic friction are .40 and .30 respectively. the box slides down unless the applied force has magnitude 13n. What is the mass of the box?
2006-10-26 23:53:05 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The box is in equilibrium if the applied force=13N.
Therefore, the total downward forces=total upward forces.
There's only 1 downward force, which is the weight of the box, W.
T here are 2 upward forces: force of friction, f, which is equal to u*R. where f is in N, u the coefficient of static friction, R the force normal to the wall in N. The other upward force is the vertical component of the applied force or 13sin28.
The box is in equilibrium. It's not moving, so use the coefficient of static friction=0.4.
R is the horizontal component of the applied force or 13cos28. Therefore,
f=0.4*13cos28.
The upward forces =13sin28+0.4*13cos28. Equate this to the downward force, W:
W=13sin28+0.4*13cos28.
= ?N.
Mass =W/g=W/9.8
=? Kg.
I'm sure you can do the rest.
2006-10-27 01:51:57 · answer #1 · answered by tul b 3 · 0⤊ 0⤋
since Mass = Force/Acceleration, and Acceleration is g=9.8 m/s2
you need to calculate the force F opposing to the force of gravity G
that's the friction Ff which is equal to the coefficient K multiplied by the perpendicular force on the surface Fp
this force Fp is the horizontal component of the total force Ft applied to the box. and you can calculate it by using a trigonometric function (sin or cos).
I don't understand why do you have both the kinetic and the static coefficients since you say it's held in place... (hence static)
hope I'm correct
2006-10-27 07:04:05 · answer #2 · answered by ╠╬╣ 3 · 0⤊ 0⤋