A block of mass weighing 200 kg is being pulled upward along an inclined surface. The angle of the surfacewith respect to the horizontal axis is 15°. The force pulling the block is equivalent to 600 N and is inclined 8° upward withrespect to the axis of the inclined surface.a) Determine the component of the pulling force in the horizontal axis.b) Determine the component of the pulling force in the axis perpendicular to the inclined surface.c) Determine the component of the self-weight of the crate in the axis parallel to the line of action of the pullingforce.

:) Two racquetballs, each having a mass of 170 g. are placed in a glass jar as shown. Their centers lie on a straight line that makes a 45 degree angle with the horizontal. a) Assume the walls are frictionless and determine P1, P2 and P3. b) Determine the magnitude of the force exerted by the left ball on the right ball. 45°

Pushing a Block and a Wedge 0 points possible (ungraded) In the following problem, you are asked to: i) Define the system and the environment: Identify which objects - the environment - interact with the object of interest - the system. Justify your choice of system: would a different choice also work? If not, why not? If so, why is the choice you made better? ii) Describe the interactions and the resulting forces: Identify the forces exerted on the object of interest by the other objects. iii) Draw the free body diagram of the object: Make sure that each force arrow is labeled with two subscripts - who is exerting the force and on whom is the force being exerted. The length of the arrows should represent qualitatively the magnitude of the force. a) c) M m 0 a F M m 0 A block of mass m rests on a frictionless wedge which has a mass M and whose surface makes an angle to the horizontal as Calculator

The figure below illustrates a hydraulic jack.a) Calculate the force F1 exerted by the operator to lift a load F2 of 104N, knowing that the diameter of the pump piston is 6 mm while the diameter of the load cylinder is equal to 25 mm. Answer: 576 N b) Determine the diameter of the load cylinder so that the force F1 is equal to 100 N. Answer: 0.06m

3) For the following system, the rod is uniform and weighs 1200N and the sign weighs 2000N. The diameter of the cylindrical supporting steel cable is 4.00cm. a) Determine the tension needed in the steel cable to keep the system in equilibrium. b) Determine the horizontal and vertical forces on the rod due to the pivot. c) Determine the fractional change in the steel cable. Supporting Cable, with tension FT 3/4L 25°/0 L Pivot x=90° A sign B = 65°

A)DETERMINE MAGNITUDE OF FORCE IN CORD HA ( IN NEWTON) B)DETERMINE MAGNITUDE OF FORCE IN CORD AB (IN NEWTON) C) DETERMINE THE MAGNITUDE OF FORCE IN CORD BD (IN NEWTON) D)DETERMINE THE MAGNITUDE OF FORCE IN CORD BC (IN NEWTON)

A tugboat with a displacement¹ of 50 tonnes and resistance² of 750 kN is towing a 75-tonne barge with resistance of 1000 kN. The tugboat has a propeller generates thrust³ of 7 to move it forward, while the barge is non-propelled. Determine the thrust and the force exerted by the rope on the tugboat, given that it takes 35 seconds for the tugboat to uniformly increase its speed to 13.5 knots4. ces: 1. The displacement or displacement tonnage of a ship is its weight. It refers to the weight or mass of water that a ship displaces when it is floating in water. 2. Ship resistance, also known as hydrodynamic resistance, refers to the force that opposes the motion of a ship or vessel as it moves through water. 3. The thrust of a propeller refers to the force generated by a rotating propeller that propels a vessel through water. 4. Knot is a unit of speed used to measure the speed of a vessel. One knot is equal to 0.5144 m/s. 2: OO TT

1.1 Preview 'em No. 2: a) Find the magnitude and location of the resultant force if a=1m. b) find the reaction forces at A and B. 1800 N/m A a 4 m- 600 N/m B

1.14 In the hoist drive system shown in Fig.22 , the mass M is considered being moved upwards with negligible frictional torque. Show that the load torque and the equivalent moment of inertia are given by dwm Tm = Je dt W2 М.r Wm Je = Im + J1 + ( [J2 + M.r²] Wm Wm Om, Tm Ji Motor J2 2r Jm 02 gear Loss-free Fig.22.

solve last three parts only