posted by Tim on .
Two 10 kilogram boxes are connected by a massless frictionless pulley. The boxes remain st rest with the one on the right hanging vertically and the one on the left 2.0 meters from the bottom of an inclined plane that makes an angle of 60 degrees with the horizontal. The coefficients of kinetic friction and static friction between the left hand box and the plane are 0.15 and 0.30 respectively. You may use g= 10 m/s(2nd power), sin 60 degrees = .087, and cos 60 degrees = 0.50.
What is the tension in the string
What is the force/ forces on the box that is on the plane
You need to break the forces into components along the direction of travel.
First, the hanging box, its weight is mg, downward, in the direction of travel.
Second, the box on the plane. The weigh mg has a component along the plane, mgsinTheta, and a component normal to the plane, mgCosTheta. Friction, then, will be mgCosTheta*mu
Next, the force equation.
assume the movement is to the right (clockwise)
so solve for a. If it is positive, your assumptions about direction (and the direction of friction) was correct.
Next tension. The force on the rope just above the hanging box is
mg-ma. The force on the box on the plane is mgSinTheta+mgCosTheta*mu + ma
check my thinking.
Your logic is terrible. Fix it.