Equilibrium practise exersises with key

EQUILIBRIUM - Practice Exercises Formulas: τ = rFsinθ Στ = 0 ΣF = 0 1. If the torque needed to loosen a lug nut holding the wheel of a car is 45 N·m and you are using a wheel wrench that is 35 cm long, what is the magnitude of the force you exert perpendicular to the end of the wrench? τ = Fd F = τ/d = 45 N·m / 0.35 m = 130 N 2. A beam of negligible mass is attached to a wall by means of a hinge. Attached to the centre of the beam is a 400 N weight. A rope also helps to support this beam as shown   below. [Image of a beam supported by a rope and a wall] a) What is the magnitude of the tension in the rope? Στcw = Στccw 400(0.5) = Ftsin40°(2) Ft = 311 N b) What are the forces that the wall exerts on the beam? Vertical: Fwall,up + Ftension,up = 400 N Fwall,up = 400 N - 311 N = 89 N Horizontal: Fwall,right = Ftension,horizontal = 311 N * sin 40° = 200 N 3. A 650 N student stands on a 250 N uniform beam that is supported by two supports as shown below. [Image of a beam supported by two supports with a person standing on it] If the supports are 5.00 m apart and the student stands 1.50 m from the left support, a) what is the magnitude of the force that the right support exerts on the beam? Στcw = Στccw (650 N)(1.5 m) + (250 N)(2.5 m) = Fr(5.0 m) Fr = 320 N Lesson 2 - Practice Exercises **3. A uniform 400 N diving board is supported at two points as shown below.** [Image of a diving board with two supports and a person standing on the end] If a 75.0 kg diver stands at the end of the board, what are the forces acting on each support? Στcw = Στccw (75 kg)(9.81 m/s²)(L) = Ft(L) + (400 N)(L/2) Ft = 3433 N = 3.43 x 10³ N F1 + F2 = 400 N + 735.75 N = 1135.75 N F2 = 1135.75 N - Ft = 792.45 N ≈ 792 N **4. Find the tension in the rope supporting the 200 N hinged uniform beam shown below.** [Image of a beam hinged to a wall and supported by a rope] Στcw = Στccw (200 N)(L/2) = (Ft sin 30°)(L) Ft = 200 N / sin 30° = 400 N **5. Find the tension in the rope supporting the 200 N hinged uniform beam shown below.** [Image of a beam hinged to a wall and supported by a rope at an angle] Στcw = Στccw (200 N)(L/2) = (Ft sin 50°)(L) Ft = 200 N / sin 50° = 261 N PRACTICE TEST 1. A box with the mass m is sliding at a constant velocity down an incline as shown. Write an expression for the magnitude of the force of friction on this box. Fnet = 0 = Fapp - Fgrav 0 = Fapp - Fgx Fapp = Fgx Fapp = mg sin θ F = μFnorm F = μmg cos θ 2. A 5.0 m long uniform beam with a mass of 8.0 kg is placed on a pivot as shown in the illustration. If a 32 kg mass is placed 1.5 m from the pivot, where should a 42 kg mass be placed on the beam to keep the beam in static equilibrium? [Image of a beam with two masses] Στcw = Στccw 32(9.81)(1.5) = 42(9.81)d d = 1.1 m 3. Which properties of a beam are balanced if a wooden beam has no rotational motion? The torques are balanced. FORCES CAUSE MOTION - Practice Test 4. A 2.55 kg box slides down a 30.0° incline at a constant velocity. What is the force of friction between the block and the incline? Fnet = 0 = Fgx - Ff Ff = Fgx = mg sin θ = (2.55 kg)(9.81 m/s²)(sin 30°) = 12.5 N 5. Two masses (m1 = 3.00 kg, m2 = 5.00 kg) hang from the ends of a metre stick as shown in the diagram. If the mass of the metre stick is negligible, at what distance from the left of the metre stick should a pivot be   placed so that the system will be balanced?   Στcw = Στccw d(3 kg)(9.81 m/s²) = (5 kg)(9.81 m/s²)(1-d) 3d = 5 - 5d 8d = 5 d = 5/8 = 0.625 m from left end of meter stick 6. A 3.0 kg block slides down a 37° incline at a constant velocity. What is the coefficient of friction between the block and the incline? Fnet = 0 = Fgx - Ff Ff = Fgx = mg sin θ μFnorm = mg sin θ μmg cos θ = mg sin θ μ = sin θ / cos θ = tan 37° = 0.75 7. In the static arrangement shown in the illustration, what is F1? F1 = m1g = (13 kg)(9.81 m/s²) = 127.53 N ≈ 1.28 x 10² N