2. A 200-lb block is at rest on a 30 degree incline. The coefficient of friction between the block and the incline is 0.20. Compute the value of the horizontal force P that will cause motion to impend up the incline.

A carton of mass 10 kg is at rest on a rough plane inclined at an angle of 25° to the horizontal. Calculate the magnitude of the friction force, in newtons, to two significant figures. (Take the magnitude of the acceleration due to gravity, g, to be 9.8 ms 2.) The magnitude of the friction force is N (to 2.s.f.).

Determine the magnitude P of the vertical force required to lift the wheelbarrow free of the ground at point B. The combined weight of the wheelbarrow and its load is 273 lb with center of gravity at G. 23" Answer: P 16" B 20" 8" lb 21"

A 60-mm-diameter solid aluminum post is subjected to a horizontal force of V = 25 KN, a vertical force of P = 70 kN, and a concentrated torque of T = 3.25 kN-m, acting in the directions as shown in the Figure. If L = 90 mm. Determine the normal and shear stresses at (a) point Hand (b) point K. T "| P HK X L

Write Newton’s Second Law of Motion for three-dimensionalmotion with only the gravitational force (acting in the z-direction).

Find a unit vector parallel to the tangent line of y = x at (2,8). What angle does this vector make with a horizontal line in the positive direction?

The graph of a force function (in newtons) is given. How much work is done by the force AF (N) 24 20 16 124 Q in moving an object 10 m? 4 x (m) 0- The amount of work done by the force in moving an object 10 m is J. (Type a whole number.) Le

please answer!!

A cable exerts a constant upward tension of magnitude 2.14 ✕ 104 N on a 2.00 ✕ 103 kg elevator as it rises through a vertical distance of 3.90 m. HINT (a) Find the work done by the tension force on the elevator (in J).  J (b) Find the work done by the force of gravity on the elevator (in J).  J

A seasoned parachutist went for a skydiving trip where he performed freefall before deploying the parachute. According to Newton's Second Law of Motion, there are two forcës acting on the body of the parachutist, the forces of gravity (F,) and drag force due to air resistance (Fa) as shown in Figure 1. Fa = -cv ITM EUTM FUTM * UTM TM Fg= -mg x(t) UTM UT UTM /IM LTM UTM UTM TUIM UTM F UT GROUND Figure 1: Force acting on body of free-fall where x(t) is the position of the parachutist from the ground at given time, t is the time of fall calculated from the start of jump, m is the parachutist's mass, g is the gravitational acceleration, v is the velocity of the fall and c is the drag coefficient. The equation for the velocity and the position is given by the equations below: EUTM PUT v(t) = mg -et/m – 1) (Eq. 1.1) x(t) = x(0) – Where x(0) = 3200 m, m = 79.8 kg, g = 9.81m/s² and c = 6.6 kg/s. It was established that the critical position to deploy the parachutes is at 762 m from the ground…