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kykrilka [37]
3 years ago
13

A 1170-kg car is held in place by a light cable on a very smooth (frictionless) ramp, as shown in the figure (Figure 1) . The ca

ble makes an angle of 31.0 ∘ above the surface of the ramp, and the ramp itself rises at 25.0 ∘ above the horizontal.
Draw a free-body diagram for the car.

Physics
1 answer:
liq [111]3 years ago
4 0
Refer to the diagram shown below.

The mass of the car is 1170 kg, therefore its weight is
W = (1170 kg)*(9.8 m/s²) = 11466 N

The component of the weight acting down the incline is
F = W sin(25°) = (11466 N)*sin(25°) = 4845.7 N

The normal reaction from the inclined plane is
N = W cos(25°) = (11466 N) cos(25°) = 1039.2 N

T =  tension in the cable, acting at 31° above the surface of the ramp.

The Free Body Diagram on the right shows all the forces (friction is ignored)
and they FDB is sufficient for determining the value of T which establishes equilibrium.

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Find the magnitude of the resultant force and the angle it makes with the positive x-axis. (Let |a| = 22 lb and |b| = 16 lb. Rou
SVEN [57.7K]

Incomplete question as the angle between the force is not given I assumed angle of 55°.The complete question is here

Two forces, a vertical force of 22 lb and another of 16 lb, act on the same object. The angle between these forces is 55°. Find the magnitude and direction angle from the positive x-axis of the resultant force that acts on the object. (Round to one decimal places.)  

Answer:

Resultant Force=33.8 lb

Angle=67.2°

Explanation:

Given data

Fa=22 lb

Fb=16 lb

Θ=55⁰

To find

(i) Resultant Force F

(ii)Angle α

Solution

First we need to represent the forces in vector form

\sqrt{x} F_{1}=22j\\ F_{2}=u+v\\F_{2}=16sin(55)i+16cos(55)j\\F_{2}=16(0.82)i+16(0.5735)j\\F_{2}=13.12i+9.176j

Total Force

F=F_{1}+F_{2}\\ F_{2}=22j+13.12i+9.176j\\F_{2}=13.12i+31.176j

The Resultant Force is given as

|F|=\sqrt{x^{2} +y^{2} }\\|F|=\sqrt{(13.12)^{2} +(31.176)^{2} }\\ |F|=33.8lb

For(ii) angle

We can find the angle bu using tanα=y/x

So

tan\alpha =\frac{31.176}{13.12}\\ \alpha =tan^{-1} (\frac{31.176}{13.12})\\\alpha =67.2^{o}

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Option B The thickness of the central portion of a thin conveying lens can be determined very accurately by using a micrometer screw gauge.

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<h3>What purposes does a micrometer serve?</h3>

A tool known as a micrometer is used to measure solid objects’ lengths, thicknesses, and other dimensions precisely and linearly.

<h3>What is the micrometer screw gauge’s SI unit?</h3>

The SI symbol m is also known as a micron, which is an SI-derived unit of length equaling 1106 meters, where 106 is the SI standard prefix for the prefix “micro-.” A micrometer is one-millionth of a meter.

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