Answers to Inclined Plane Problems 1. A box of mass 25 kg is placed on an inclined plane that is at a 40o angle with the horizontal. m = 25 kg (a) Draw a sketch. 405o Fperp 40o Wt. Fpar
1. A box of mass 25 kg is placed on an inclined plane that is at a 40o angle with the horizontal. (b) Determine the weight of the box. m = 25 kg Wt. = m g = ( 25 kg )( 9.8 m/s2 ) 40o Fperp 40o Wt. Wt. = 245 N Fpar
1. A box of mass 25 kg is placed on an inclined plane that is at a 40o angle with the horizontal. (c) Determine the normal force FN and the parallel force Fp. m = 25 kg FN = Fperp 40o Fperp 40o 245 N Fpar Fpar Fperp sin 40 = cos 40 = 245 245 Fpar = 245 sin 40 Fperp = 245 cos 40 Fperp = 188 N Fpar = 157 N FN = 188 N
1. A box of mass 10 kg is placed on an inclined plane that is at a 25o angle with the horizontal. (d) Assume the plane is frictionless. Find the acceleration of the block down the plane. m = 25 kg a = ? 40o 188 N 40o F = m a 245 N F F = Fpar 157 N a = m 157 N = 25 kg a = 6.3 m/s2
2. A 1500-kg car is on an inclined slope at an angle of 20o with the horizontal. It has been left in neutral without the parking brake on. (a) Determine FN and Fp. m = 1500 kg Wt. = m g = ( 1500 kg )( 9.8 m/s2 ) Fperp 20o 20o Wt. = 14 700 N 14 700 N Fpar Fperp Fpar cos 20 = sin 20 = 14 700 14 700 Fperp = 14 700 cos 20 Fpar = 14 700 sin 20 Fperp = 13 813 N Fpar = 5028 N FN = 13 813 N
2. A 1500-kg car is on an inclined slope at an angle of 20o with the horizontal. It has been left in neutral without the parking brake on. (b) If the coefficient of friction is 0.35, calculate the force of friction. m = 1500 kg Ff = μ FN 13 813 N 20o 20o 14 700 N = ( 0.35 )( 130813 N ) 5028 N Ff = 4834 N
2. A 1500-kg car is on an inclined slope at an angle of 20o with the horizontal. It has been left in neutral without the parking brake on. (c) Will the car begin to slide down the slope? m = 1000 kg 4834 N 13 813 N 20o 20o 14 700 N 5028 N Fp > Ff , so car begins to slide down the slope
3. A crate of mass 80 kg is placed on an inclined plane at an angle of 40o with the horizontal. The coefficient of friction between the crate and the plane is 0.20. Find the acceleration of the crate down the plane. Wt. = m g m = 80 kg = ( 80 kg )( 9.8 m/s2 ) a = ? Wt. = 784 N Fperp 40o 40o Wt. Fpar
3. A crate of mass 80 kg is placed on an inclined plane at an angle of 40o with the horizontal. The coefficient of friction between the crate and the plane is 0.20. Find the acceleration of the crate down the plane. Wt. = m g m = 80 kg = ( 80 kg )( 9.8 m/s2 ) a = ? Wt. = 784 N Fperp 40o 40o 784 N Fperp cos 40 = Fpar Fpar 784 sin 40 = 784 Fperp = 784 cos 40 Fpar = 784 sin 40 Fperp = 601 N Fpar = 504 N FN = 601 N
3. A crate of mass 80 kg is placed on an inclined plane at an angle of 40o with the horizontal. The coefficient of friction between the crate and the plane is 0.20. Find the acceleration of the crate down the plane. m = 80 kg Ff Ff = μ FN a = ? = ( 0.20 )( 601 N ) 601 N Ff = 120 N 40o 40o 784 N 504 N
3. A crate of mass 80 kg is placed on an inclined plane at an angle of 40o with the horizontal. The coefficient of friction between the crate and the plane is 0.20. Find the acceleration of the crate down the plane. m = 80 kg 120 N Fnet = Fpar + Ff a = ? = ( +504 N ) + ( - 120 N ) Fnet 601 N 40o Fnet = 384 N 40o 784 N F F = m a a = 504 N m 384 N = 80 kg a = 4.8 m/s2
4. A man wishes to push a refrigerator up a ramp that is at an angle of 15o with the ground. The refrigerator weighs 1000 N. The coefficient of friction is 0.14. (a) Determine the components FN and Fp. 15o Fperp 15o 1000 N Fperp Fpar Fpar cos 15 = sin 15 = 1000 1000 Fperp = 1000 cos 15 Fpar = 1000 sin 15 Fperp = 966 N Fpar = 259 N FN = 966 N
4. A man wishes to push a refrigerator up a ramp that is at an angle of 15o with the ground. The refrigerator weighs 1000 N. The coefficient of friction is 0.14. (a) Determine the components FN and Fp. (b) Determine the force of friction. 15o 966 N Ff = μ FN 15o 1000 N = ( 0.14 )( 966 N ) 259 N Ff = 135 N
4. A man wishes to push a refrigerator up a ramp that is at an angle of 15o with the ground. The refrigerator weighs 1000 N. The coefficient of friction is 0.14. (a) Determine the components FN and Fp. (b) Determine the force of friction. (c) What force must the man exert to push the safe up the ramp? FA = ? 135 N 15o 966 N 15o 1000 N Man must exert enough force to overcome both Fpar and Ff 259 N so FA = ( 259 N ) + ( 135 N ) FA = 394 N