Force problem

  • Thread starter phizuks
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  • #1
phizuks
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I'm not sure why I'm not getting this problem right. It seems pretty simple to me, but I guess there might be a trick to it?

Suppose that a 50kg block slides along a horizontal surface where the coefficient of kinetic friction between the block and the surface is uk = 0.60. A force F = 400 N is now applied as shown in the drawing, where the angle of the force above horizontal is 20°.

What is the magnitude of the acceleration of the block?

Isn't the equation just 400N*Cos(20)-(9.81 m/s^2)(50kg)(.60) = (50kg)(a)

I get a = 1.63 m/s^2
The choices the online homework gives are
.54, 2.31, 3.27 (2x the accel i get), 6.78, and 8.11
 

Answers and Replies

  • #2
Päällikkö
Homework Helper
519
11
You need to take into account the increased N, as it now has to deal with both gravity and the y-component of the F. This, of course, increases the force caused by friction.
 
  • #3
phy_bits
5
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phizuks said:
I'm not sure why I'm not getting this problem right. It seems pretty simple to me, but I guess there might be a trick to it?

Suppose that a 50kg block slides along a horizontal surface where the coefficient of kinetic friction between the block and the surface is uk = 0.60. A force F = 400 N is now applied as shown in the drawing, where the angle of the force above horizontal is 20°.

What is the magnitude of the acceleration of the block?

Isn't the equation just 400N*Cos(20)-(9.81 m/s^2)(50kg)(.60) = (50kg)(a)

I get a = 1.63 m/s^2
The choices the online homework gives are
.54, 2.31, 3.27 (2x the accel i get), 6.78, and 8.11



the normal reaction here will be less than 50*9.8...bcoz the force F has a component in the vertical dir.
N=50*9.8 - 400*sin(20)

so ur eqn shud actually b
400N*Cos(20)-[(9.81 m/s^2)(50kg)-(400)(sin(20))](.60) = (50kg)(a)

i think u shud get the ans frm this.

bye
 

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