3.014 m/s (b) Find the acceleration of the block if the coefficlent of kinetic f
ID: 1656721 • Letter: 3
Question
3.014 m/s (b) Find the acceleration of the block if the coefficlent of kinetic friction between the block and incline is 0.11. (Give the magnitude of the acceleration.) 2.03 m/s2 Need Help? Eien 12.5 points Ser P11 4P027 A donkey is harnessed to a sled having a mass of 201 kg, including supplies. The donkey must exert a force exceeding 1230 N at an angie of 30.3 (above the horizontal) in order to get the sled moving. Treat the sled as a point particle. Notes Ask Your Teach (a) Calculate the normal force (in N) on the sled when the magnitude of the applied force is 1230 N. (Enter the magnitude.) (b) Find the coefficient of static friction between the sled and the ground beneath it. the static friction force (in N) when the donkey is exerting a force of 6.15 x 10? N on the sled at the same angle. Enter the magnitude.) My Notes Ask Your A 180.2-N bind feeder is supported by three cables as shown in the figure below. Find the tension in each cable. left cable right cable9O bottom cable 80.2Explanation / Answer
(a) Weight of the sled, W = mg = 201*9.81 = 1971.8 N
Horizontal component of the force exerted by the donkey, Fx= 1230*cos30.3 = 1062 N
Vertical component of the force exerted by the donkey, Fy= 1230*sin30.3 = 621 N
So, normal force on the sled, R = W - Fy = 1971.8 - 621 = 1350.8 N
(b) Suppose k is the coefficient of static friction.
So, in the equilibrium condition -
Fx = k*R = k*1350.8
=> k = 1062 / 1350.8 = 0.79
(c) Normal force in this case, R = W - Fy = 1971.8 - 615*sin30.3 = 1661.5 N
So, frictional force, Ff = k*R = 0.79*1661.5 = 1312.6 N
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