1) Now a simple resolution of forces question. in the figure below, a Force is s

Question

1) Now a simple resolution of forces question. in the figure below, a Force is shown by the length l = 12.6Newtons, and the angle shown a = 55.0 degrees.

Find the horizontal component of the force Fx in Newtons
Give your answer to 2 decimal places

2) A body of mass = 115.2 kg sits on a flat surface which is at an angle b = 5.1 degrees to the horizontal. What is the force F1 acting down the slope (in Newtons)

3) A body of mass = 385.0 kg sits on a flat surface which is at an angle c = 77.3 degrees to the vertical, as shown below.
What is the force F2 acting down the slope (in Newtons)

4) A body of mass = 288 kg sits on a flat friction free surface which is at an angle d = 12 degrees to the horizontal. What is the acceleration of the body down the slope (in m/s2)

5) A body of mass = 423.7 kg sits on a flat friction free surface which is at an angle e = 81.6 degrees to the vertical, as shown below. If a braking force of 200 Newtons is applied to the body, what is the acceleration down the slope in m/s2?
(Note if acceleration down the slope proves to be a negative value, include this in your answer)

6)Two bodies, each of mass = 186.7 kg sit on a flat friction free surface which is at an angle f = 12.8 degrees to the horizontal. The two objects are joined by a solid link of negligible mass.
What is the acceleration of the two bodies down the slope (in m/s2)

7)Two bodies have the properties, mass of m1 = 352.0 kg and m2 = 500 kg. They sit on a flat friction free surface which is at an angle g = 9.6 degrees to the horizontal. The two objects are joined by a solid link of negligible mass.

Explanation / Answer

1. Fx = FCos55 = 12.6*cos55 = 7.23 N

2. F = 115.2 *9.81*Sin5.1 = 100.46 N

3. F = 385*9.81*cos77.3 = 830.33 N

4. gSin12 = 9.81*sin12 = 2.039 m/s^2

5. Deceleration due to braking = F/m = 200 / 423.7 = 0.472 m/s^2

Net acceleration = gcos81.6 - 0.472 = 9.81*cos81.6 - 0.472 = 0.961 m/s^2

6.

9.81*sin12.8 = 2.173 m/s^2

7.

Force in coupler = m1*g*sin theta + m2*g*sin theta - Braking force = (m1+m2)*g*sin theta - B

= (352 + 500)*9.81*sin9.6 - 390

= 1003.87 N

8.

angle of force with incine = 20+13.4 = 33.4 deg

acceleration alon the slope due to force = (F/m)cos33.4 = (200 / 130.7)*cos33.4 = 1.278 m/s^2

Acceleration due to gravity = gSin13.4 = 9.81*sin13.4 = 2.273 m/s^2

Total acceleration = 2.273 + 1.278 = 3.551 m/s^2

9.

Figure is missing. Please provide.

10.

Figure is missing. Pl provide.

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