In order to move an 800.0 kg pallet load of bricks down a smooth concrete drivew
ID: 1273314 • Letter: I
Question
In order to move an 800.0 kg pallet load of bricks down a smooth concrete driveway you tie a rope between the pallet at ground level and the tow bar on the back of your car (the rope makes an angle of 35.0 relative to the road). The coefficient of static friction between the pallet and the driveway is p, = 0.180. If you gently start pulling on the rope using the car, how much tension is there in the rope when the pallet just starts moving (i.e. what is the magnitude of the tension force)? (Hint: draw a free body diagram of all the forces involved, take the acceleration due to gravity (= 9.81 ms-2.) (1530 N)Explanation / Answer
let T be the tension in rope
now horizontal component provided by rope = T cos35 (as 35 is the angle between road and rope)
vertical component = Tsin35
in order to just move the pallet horizontal force should be equal to friction force
friction force = uN = u (mg - Tsin35) = 0.18(800*9.81 - Tsin35)
horizontal force = Tcos35
solving we get T = umg/(usin35 + cos35) = 1531.49 N