Solving for Acceleration: 8.3kg Block & 25.3° Angle

AI Thread Summary
To determine the acceleration of an 8.3 kg block being pulled by a 53.0 N force at a 25.3° angle, the applied force must be resolved into horizontal and vertical components using trigonometric functions. The horizontal component can be calculated as F_x = 53.0 N * cos(25.3°), while the vertical component is F_y = 53.0 N * sin(25.3°). The net force acting on the block will influence its acceleration, which can be found using Newton's second law, F = ma. The discussion highlights the importance of correctly identifying the angle in calculations, as initial confusion arose with a reference to 30 degrees instead of the correct angle. Understanding these components is crucial for accurately solving for the block's acceleration.
chaotixmonjuish
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A block of mass 8.3 kg is pulled along a horizontal frictionless floor by a cord that exerts a force of 53.0 N at an angle 25.3° above the horizontal. What is the magnitude of the acceleration of the block?

I'm just wondering how I set up an equation with an angle.
 
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Separate the applied force into its horizontal and vertical components using the trigonometric ratios or use the method of tension co-efficients.
 
That's where I'm having a problem.

would it be something like:

y=(mass)(sin(30)a)
x=(mass)(cos(30)a)
 
How about F_x = F(cos (\frac{\pi}{6})) ?
 
wow, how did you get that
 
chaotixmonjuish said:
That's where I'm having a problem.

would it be something like:

y=(mass)(sin(30)a)
x=(mass)(cos(30)a)
Yes, something like that. But of course the angle isn't 30 degrees.
 

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