What force is needed to push (at constant speed) a 20 kg crate up a 32 degree in
ID: 1539583 • Letter: W
Question
What force is needed to push (at constant speed) a 20 kg crate up a 32 degree incline if the coefficient of kinetic friction between crate and incline is 0.20? A 0.35 N force is applied to a 2.2 kg object, initially at rest. The resulting acceleration is In the figure shown, m_A = 1.5 kg, m_B = 0.50 kg, and the coefficient of kinetic friction between m_B and the horizontal surface is 0.22. Find the acceleration of the system if the pulley is frictionless. A 50 kg box rests on a 35 degree incline. What is the smallest coefficient of static friction between the box and the surface of the incline for the box to remain at rest? Find the speed of a 2.0 kg mass undergoing circular motion with radius 10 cm caused by a 40 N force. In the figure shown, M = 100 g, alpha = 42 degree and beta = 55 degree. Find the tension T in the right side of the figure. Find the gravitational force of attraction between two 100 kg human beings whose centers of mass are separated by 60 cm.Explanation / Answer
4)
up the incline
Fnet = F - Fg - fk
from newtons second law Fnet = m*a = 0
F - Fg - fk = 0
F = Fg + fk
F = m*g*sintheta + uk*m*g*costheta
F = (20*9.8*sin32) + (0.2*20*9.8*cos32)
F = 137.1 N <<<<<<-------answer
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5)
from newtons second law
F = m*a
acceleration a = F/m
a = 0.35/2.2 = 0.16 m/s^2 <<<---answer
================
6)
for mB
T - fk = mB*a
for mA
mA*g - T = mA*a
mA*g - fk = a*(mA+mB)
mA*g - uk*mB*g = a*(mA+mB)
a = ((1.5*9.8) - (0.22*0.5*9.81))/(1.5+0.5)
acceleration a= 6.8 m/s^2
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)7)
In equilibrium
Fnet =0
Fg - fs = 0
m*g*sintheta - us*m*g*costheta = 0
us = tantheta = tan35 = 0.7
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8)
IN ircular motion
centripetal force Fc = mv^2/r
40 = 2*v^2/0.1
v = 1.414 m/s
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9)
along horizontal
TL*cosalpha = Tr*cosbeta
TL*cos42 = Tr*cos55
TL = Tr*0.772
along vertical
TL*sinalpha + Tr*sinbeta = Mg
(Tr*0.772*sin42) + (Tr*sin55) = 0.1*9.8
Tr = 0.734 N <<<<<-----answer
======================
F = G*m1*m2/r^2
F = 6.67*10^-11*100*100/0.6^2
F = 1.85*10^-6 N <<<-----answer
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