- #1

danago

Gold Member

- 1,122

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Hey. In physics, we are studying motion, and the three laws of motion.

At the moment im having trouble with inclined planes. Take the following example for instance:

"A frictionless plane is inclined at 50 degrees to the horizontal. A mass of 65kg is released on the slope. What is its acceleration down the slope?"

From what i understand, the weight/gravity force should be resolved into its horizontal and vertical components relative to the slope (slope being horizontal). In this case, the weight force becomes:

[tex]

W = \left( {\begin{array}{*{20}c}

{65g\cos 40} \\

{ - 65g\sin 40} \\

\end{array}} \right)

[/tex]

To balance the vertical component, so the object doesnt move up and down, the normal force therefore has the same magnitude but opposite direction as the vertical component of the weight force. Therefore:

[tex]

N = \left( {\begin{array}{*{20}c}

0 \\

{65g\sin 40} \\

\end{array}} \right)

[/tex]

The net force then ends up being the horizontal component of the weight/gravity force...[tex]

\sum F={65g\cos 40}[/tex]

Since i want to find acceleration, i then divide the force being applied to the object by its mass, 65kg. So the acceleration ends up being [tex]a={g\cos 40}=7.5 m/s/s[/tex]

However, apparently the answer is like 12.8 m/s/s. My teacher says that i need to create a right angled triangle from the information, where 9.8 m/s/s (gravitational acceleration) is the side opposite to the 50 degree angle, and then solve for the hypotenuse, which should give me the actual acceleration i need.

Could someone please tell me where ive gone wrong in my calculations. Thanks,

Dan.

At the moment im having trouble with inclined planes. Take the following example for instance:

"A frictionless plane is inclined at 50 degrees to the horizontal. A mass of 65kg is released on the slope. What is its acceleration down the slope?"

From what i understand, the weight/gravity force should be resolved into its horizontal and vertical components relative to the slope (slope being horizontal). In this case, the weight force becomes:

[tex]

W = \left( {\begin{array}{*{20}c}

{65g\cos 40} \\

{ - 65g\sin 40} \\

\end{array}} \right)

[/tex]

To balance the vertical component, so the object doesnt move up and down, the normal force therefore has the same magnitude but opposite direction as the vertical component of the weight force. Therefore:

[tex]

N = \left( {\begin{array}{*{20}c}

0 \\

{65g\sin 40} \\

\end{array}} \right)

[/tex]

The net force then ends up being the horizontal component of the weight/gravity force...[tex]

\sum F={65g\cos 40}[/tex]

Since i want to find acceleration, i then divide the force being applied to the object by its mass, 65kg. So the acceleration ends up being [tex]a={g\cos 40}=7.5 m/s/s[/tex]

However, apparently the answer is like 12.8 m/s/s. My teacher says that i need to create a right angled triangle from the information, where 9.8 m/s/s (gravitational acceleration) is the side opposite to the 50 degree angle, and then solve for the hypotenuse, which should give me the actual acceleration i need.

Could someone please tell me where ive gone wrong in my calculations. Thanks,

Dan.

Last edited: