Answer to Question #37261 in Mechanics | Relativity for alzahra
2013-11-22T09:27:31-05:00
A 282-kg crate is being pushed across a horizontal floor by a force P that makes an angle of 48.1 ° below the horizontal. The coefficient of kinetic friction is 0.288. What should be the magnitude of P, so that the net work done by it and the kinetic frictional force is zero?
http://edugen.wileyplus.com/edugen/courses/crs3976/art/qb/qu/c06/EAT_12259100087480_6454488344216817.gif
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2013-12-05T09:03:20-0500
Solution: m=282kg Theta=48.1 ° k=0.288 Force P has 2 components : vertical ( P_vert) and Horizontal (P_hor) P_vert=P*sin(Theta) P_hor=P*cos(Theta) ForceFriction=k*N N- reaction force from the surface N=Weight+P_vert=m*g+P_vert=m*g+P*sin(Theta) so ForceFriction=k*(m*g+P*sin(Theta)) WorkFriction=-ForceFriction*Distance ("-" because ForceFriction and Distance are the opposite direction) WorkP - work of the force P WorkP=P_hor*Distance (horizontal projection of the "P" times Distance) now , the sum of the works is zero WorkFriction+WorkP=0 -ForceFriction*Distance + P_hor*Distance=0 ForceFriction=P_hor k*(m*g+P*sin(Theta))=P*cos(Theta) k*m*g+k*P*sin(Theta)=P*cos(Theta) k*m*g=P*cos(Theta)-k*P*sin(Theta) k*m*g=P*(cos(Theta)-k*sin(Theta)) P=k*m*g/(cos(Theta)-k*sin(Theta)) cos(Theta)=cos (48.1 °)= 0.668 sin(Theta)=sin(48.1 °)=0.744 P=0.288*282*9.8/(0.668-0.288*0.744)=795.92/0.454=1753N ANSWER P=1753N
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