A 1060 kg is being pulled up a frictionless ramp at a constant speed, as shown i

Question

A 1060 kg is being pulled up a frictionless ramp at a constant speed, as shown in the figure (Figure The cable makes an angle of 31.00 above the surface of the ramp, and the ramp itself rises at 25.00 above the horizontal. Figure 1 of 1 C 31.00 Part A Draw a free-body diagram for the car. Draw all the required force vectors with their tails starting at the black dot on the frontbumper. The location and orientation of your vectors will be graded. The exact length of your vectors will not be graded. add olement vector sum x element attributes delate U reset help Fariat Thrust force CL31.0 25.00

Explanation / Answer

a)

its basically just Weight (pointing down from the front bumper, 270 degrees, Normal Force which is perpendicular to

the ramp, 115 degree, and Tension which is the rope pulling the car. 56 degrees

b)

Weight component pushing the car down the ramp = 1060 × 9.81 × sin(25) = 4394.64 N

Therefore upward component force exerted parallel to ramp by cable = 4394.64 N

Therefore tension is cable = 4394.64/cos(31) = 5126.93 N.

c)

The component of the car's weight perpendicular to the ramp pushes on the ramp. This is

1060 × 9.81 × cos(25) = 9424.33 N

The component of the tension in the cable perpendicular to the ramp lifts the car. This is:

5126.93 × sin(31) = 2640.56 N

Net force pushing down = 9424.33 - 2640.56 = 6783.77 N

which is equal to the force which the ramp exerts on the car.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.