Question

You are pushing a 50 kg lawn mower with a force of 80 N at an angle of 45 degreesfrom the ground (the force points at 45 degreesinto the ground, like how you’d expect when you are pushing a lawn mower). What are the A) horizontal and B) vertical component of this force? The ground is pushing straight upwards at the lawn mower with some amount of normal force. The grassy lawn is also exerting a friction force against the motion of the lawn mower. As a result of the normal force and friction force, the lawn mower is moving at constant speed (zero acceleration, zero net force). C) Then, what is that normal force exerted by the ground? D) What is that friction force exerted by the ground? (round all answers to integer) (up is + down is -) (use g=9.81m/s²)

A) 57

B)-57

C)?

D)?

Answer 1

=90N Fapplied 450 § 5 ing Horizontal component of applied fare: - 90 cosaso =4052N = a cos 45° - 56.56854B N + - +56.57 N (6) - TAN Vertical component of applied force B Farinero - 4052n. -156-57 N (-ŷj -FS7n ☺) こ & fyzo AN = fa sinaro + ng 4052 + sox 9.81 547.0685 425 N 1347 N = If Z { Fuso (because no acceleration & but moving 7 f = fa cosas f. = 4052 N (1) f = -56-57 N f =-57 N رثا Friction facce = -S7N.