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F net x = w x f .

Substituting this into Newton’s second law, a x = F net x / m , gives

a x = F net x m = w x f m = m g sin 25 ° f m .

We substitute known values to obtain

a x = ( 60.0 kg ) ( 9.80 m/s 2 ) ( 0.4226 ) 45.0 N 60.0 kg .

This gives us

a x = 3.39 m/s 2 ,

which is the acceleration parallel to the incline when there is 45.0 N of opposing friction.

Significance

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. It is a general result that if friction on an incline is negligible, then the acceleration down the incline is a = g sin θ , regardless of mass. As discussed previously, all objects fall with the same acceleration in the absence of air resistance. Similarly, all objects, regardless of mass, slide down a frictionless incline with the same acceleration (if the angle is the same).

When an object rests on an incline that makes an angle θ with the horizontal, the force of gravity acting on the object is divided into two components: a force acting perpendicular to the plane, w y , and a force acting parallel to the plane, w x ( [link] ). The normal force N is typically equal in magnitude and opposite in direction to the perpendicular component of the weight w y . The force acting parallel to the plane, w x , causes the object to accelerate down the incline.

Figure shows a point object on a slope of angle theta with the horizontal. Force w points vertically down from the point. Wx points down and parallel to the slope. Wy points down and perpendicular to the slope. The angle between w and wy is theta. The figure includes these equations: wx is equal to w sine theta is equal to mg sine theta, and wy is equal to w cos theta is equal to mg cos theta.
An object rests on an incline that makes an angle θ with the horizontal.

Be careful when resolving the weight of the object into components. If the incline is at an angle θ to the horizontal, then the magnitudes of the weight components are

w x = w sin θ = m g sin θ

and

w y = w cos θ = m g cos θ .

We use the second equation to write the normal force experienced by an object resting on an inclined plane:

N = m g cos θ .

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, we draw the right angle formed by the three weight vectors. The angle θ of the incline is the same as the angle formed between w and w y . Knowing this property, we can use trigonometry to determine the magnitude of the weight components:

cos θ = w y w , w y = w cos θ = m g sin θ sin θ = w x w , w x = w sin θ = m g sin θ .

Check Your Understanding A force of 1150 N acts parallel to a ramp to push a 250-kg gun safe into a moving van. The ramp is frictionless and inclined at 17 ° . (a) What is the acceleration of the safe up the ramp? (b) If we consider friction in this problem, with a friction force of 120 N, what is the acceleration of the safe?

a. 1.7 m/s 2 ; b. 1.3 m/s 2

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Tension

A tension    is a force along the length of a medium; in particular, it is a pulling force that acts along a stretched flexible connector, such as a rope or cable. The word “tension” comes from a Latin word meaning “to stretch.” Not coincidentally, the flexible cords that carry muscle forces to other parts of the body are called tendons .

Any flexible connector, such as a string, rope, chain, wire, or cable, can only exert a pull parallel to its length; thus, a force carried by a flexible connector is a tension with a direction parallel to the connector. Tension is a pull in a connector. Consider the phrase: “You can’t push a rope.” Instead, tension force pulls outward along the two ends of a rope.

Practice Key Terms 3

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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