Question
A block of mass 300kg is just pulled along rough horizontal ground by two equal force inclined at 30 degrees to the line of motion.if the coefficient of friction is 0.6.find the value of P
Answers
saiko
the normal force on the block,
mg-p*sin_theta
along the motion,
p*cos_theta=u*(mg-p*sin_theta)
theta=30
u=.6
m=300 kg
p*(.866+.5*.6)=.6*300*9.8
p=1512 N
if p is individual of two force then,
p=1512/2=756 N
mg-p*sin_theta
along the motion,
p*cos_theta=u*(mg-p*sin_theta)
theta=30
u=.6
m=300 kg
p*(.866+.5*.6)=.6*300*9.8
p=1512 N
if p is individual of two force then,
p=1512/2=756 N