Forces on an object, Force vector

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SUMMARY

The discussion focuses on calculating the second force acting on a 1.1 kg box, given its acceleration and one known force. The net force is determined using Newton's second law, F=ma, resulting in a net force of 13.2 N. The components of the net force are resolved into unit-vector notation, leading to the equation F1 + F2 = F(net), where F1 is the known force of 20 N. Participants clarify that the goal is to find the second force, F2, by resolving the net force into its components.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of vector resolution in physics
  • Familiarity with trigonometric functions (sine and cosine)
  • Ability to interpret force diagrams
NEXT STEPS
  • Study vector resolution techniques in physics
  • Learn how to apply trigonometric functions to force calculations
  • Explore examples of net force calculations in two-dimensional systems
  • Review unit-vector notation and its applications in physics
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Students studying physics, particularly those focusing on mechanics and force analysis, as well as educators seeking to enhance their teaching methods in vector resolution and force calculations.

Rileyss123
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There are two forces on the 1.1 kg box in the overhead view of Fig. 5-37 but only one is shown. The figure also shows the acceleration of the box.

IMAGE : http://www.webassign.net/hrw/05_37.gif

(a) Find the second force in unit-vector notation.
N i + N j

(b) Find the second force as a magnitude and direction.
N,
° (counterclockwise from the +x-axis is positive)

I'm not exactly sure where to start. Using F=ma then F= (1.1)(12)=13.2
so is the net Force 13.2
and for N in part be would the answer be either 20tan(30) or 20/tan(30) ??
 
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Rileyss123 said:
There are two forces on the 1.1 kg box in the overhead view of Fig. 5-37 but only one is shown. The figure also shows the acceleration of the box.

IMAGE : http://www.webassign.net/hrw/05_37.gif

(a) Find the second force in unit-vector notation.
N i + N j

(b) Find the second force as a magnitude and direction.
N,
° (counterclockwise from the +x-axis is positive)

I'm not exactly sure where to start. Using F=ma then F= (1.1)(12)=13.2
so is the net Force 13.2
and for N in part be would the answer be either 20tan(30) or 20/tan(30) ??

You are asked to find the second force and not the net force.
As you have calculated the net force and you know its direction, resolve it into components
F(net)= -13.2 sin 30 i + -13.2 cos 30 j
F1= 20 i
F1+F2=F(net)

Find F2 now
 

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