Question

Jason works for a moving company. A 45 kgwooden crate is sitting on the wooden ramp of his truck; the ramp is angled at 11∘.

Part A: What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving up the ramp? Use appropriate units.

Part B: What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving down the ramp? Use appropriate units.

Answer 1

Draw the free-body diagram for the wooden crate sitting on the wooden ram as follows, Part A The normal force on the wooden crate is The equation for the static frictional force on the crate is The coefficient of static frictional value between two wooden surfaces is μ, 0.5 The net force on the crate when Jason apply the force directed parallel to the ramp, that Jason needs to exert on the crate to get it to start moving up the ramp is Rewrite the above equation for F and substitute the values in the equation. F-mg (14,cosB+ sin θ) (45 kg)(9.81 m/s(0.5) cosl 1+sinl1) (45 kg)(9.81 m/s050.9816) (0.1908) 300.89 N Therefore, the required force to move the crate up the ramp is 300.89 N Part B, The net force on the crate when Jason apply the force parallel to the ramp, that Jason needs to exert on the crate to get it to start moving down ther amp is, f-mgsin8-F Rewrite the above equation for F and substitute the values in the equation. F-mg (14,cose-sine) (45 kg)(9.81 m/s(0.5) cosl 1-sinl) - (45 kg)(9.81 m/s') (0.5) (0.9816)-(0.1908) 132.43 N Therefore, the required force to move the crate down the ramp is 132.43 N.