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F net = w f size 12{F rSub { size 8{"net " \lline \lline } } =w rSub { size 8{ \lline \lline } } - f} {} ,

and substituting this into Newton’s second law, a = F net m size 12{a rSub { size 8{ \lline \lline } } = { {F rSub { size 8{"net " \lline \lline } } } over {m} } } {} , gives

a = F net m = w f m = mg sin ( 25º ) f m size 12{a rSub { size 8{ \lline \lline } } = { {F rSub { size 8{"net " \lline \lline } } } over {m} } = { {w rSub { size 8{ \lline \lline } } - f} over {m} } = { { ital "mg""sin" \( "25"° \) - f} over {m} } } {} .

We substitute known values to obtain

a = ( 60 . 0 kg ) ( 9 . 80 m/s 2 ) ( 0 . 4226 ) 45 . 0 N 60 . 0 kg size 12{a rSub { size 8{ \lline \lline } } = { { \( "60" "." "0 kg" \) \( 9 "." "80 m/s" rSup { size 8{2} } \) \( 0 "." "4226" \) - "45" "." "0 N"} over {"60" "." "0 kg"} } } {} ,

which yields

a = 3 . 39 m/s 2 size 12{a rSub { size 8{ \lline \lline } } =3 "." "39 m/s" rSup { size 8{2} } } {} ,

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

Discussion

Since friction always opposes motion between surfaces, the acceleration is smaller when there is friction than when there is none. In fact, it is a general result that if friction on an incline is negligible, then the acceleration down the incline is a = g sin θ size 12{a=g"sin"θ} {} , regardless of mass . This is related to the previously discussed fact that 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).

Resolving weight into components

Vector arrow W for weight is acting downward. It is resolved into components that are parallel and perpendicular to a surface that has a slope at angle theta to the horizontal. The coordinate direction x is labeled parallel to the sloped surface, with positive x pointing uphill. The coordinate direction y is labeled perpendicular to the sloped surface, with positive y pointing up from the surface. The components of w are w parallel, represented by an arrow pointing downhill along the sloped surface, and w perpendicular, represented by an arrow pointing into the sloped surface. W parallel is equal to w sine theta, which is equal to m g sine theta. W perpendicular is equal to w cosine theta, which is equal to m g cosine theta.
An object rests on an incline that makes an angle θ with the horizontal.

When an object rests on an incline that makes an angle θ size 12{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 size 12{w rSub { size 8{ ortho } } } {} , and a force acting parallel to the plane, w size 12{w rSub { size 8{ \lline \lline } } } {} . The perpendicular force of weight, w size 12{w rSub { size 8{ ortho } } } {} , is typically equal in magnitude and opposite in direction to the normal force, N size 12{N} {} . The force acting parallel to the plane, w size 12{w rSub { size 8{ \lline \lline } } } {} , causes the object to accelerate down the incline. The force of friction, f size 12{f} {} , opposes the motion of the object, so it acts upward along the plane.

It is important to be careful when resolving the weight of the object into components. If the angle of the incline is at an angle θ size 12{θ} {} to the horizontal, then the magnitudes of the weight components are

w = w sin ( θ ) = mg sin ( θ ) size 12{w rSub { size 8{ \lline \lline } } =w"sin" \( θ \) = ital "mg""sin" \( θ \) " "} {}

and

w = w cos ( θ ) = mg cos ( θ ) size 12{w rSub { size 8{ ortho } } =w"cos" \( θ \) = ital "mg""cos" \( θ \) } {} .

Instead of memorizing these equations, it is helpful to be able to determine them from reason. To do this, draw the right triangle formed by the three weight vectors. Notice that the angle θ size 12{θ} {} of the incline is the same as the angle formed between w size 12{w} {} and w size 12{w rSub { size 8{ ortho } } } {} . Knowing this property, you can use trigonometry to determine the magnitude of the weight components:

cos ( θ ) = w w w = w cos ( θ ) = mg cos ( θ ) alignl { stack { size 12{"cos" \( θ \) = { {w rSub { size 8{ ortho } } } over {w} } } {} #w rSub { size 8{ ortho } } =w"cos" \( θ \) = ital "mg""cos" \( θ \) {} } } {}

sin ( θ ) = w w w = w sin ( θ ) = mg sin ( θ ) alignl { stack { size 12{"sin" \( θ \) = { {w rSub { size 8{ \lline \lline } } } over {w} } } {} #w rSub { size 8{ \lline \lline } } =w"sin" \( θ \) = ital "mg""sin" \( θ \) {} } } {}

Tension

A tension     is a force along the length of a medium, especially a force carried by a flexible medium, 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 exert pulls only parallel to its length; thus, a force carried by a flexible connector is a tension with direction parallel to the connector. It is important to understand that tension is a pull in a connector. In contrast, consider the phrase: “You can’t push a rope.” The tension force pulls outward along the two ends of a rope.

Consider a person holding a mass on a rope as shown in [link] .

An object of mass m is attached to a rope and a person is holding the rope. A weight vector W points downward starting from the lower point of the mass. A tension vector T is shown by an arrow pointing upward initiating from the hook where the mass and rope are joined, and a third vector, also T, is shown by an arrow pointing downward initiating from the hand of the person.
When a perfectly flexible connector (one requiring no force to bend it) such as this rope transmits a force T size 12{T} {} , that force must be parallel to the length of the rope, as shown. The pull such a flexible connector exerts is a tension. Note that the rope pulls with equal force but in opposite directions on the hand and the supported mass (neglecting the weight of the rope). This is an example of Newton’s third law. The rope is the medium that carries the equal and opposite forces between the two objects. The tension anywhere in the rope between the hand and the mass is equal. Once you have determined the tension in one location, you have determined the tension at all locations along the rope.
Practice Key Terms 3

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Source:  OpenStax, Newton's laws. OpenStax CNX. Oct 25, 2015 Download for free at https://legacy.cnx.org/content/col11898/1.1
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