# Two forces of 60N and 80N acting at an angle of 60degrees with eath other. Pull an object. What single pull will replace the given forces. What is the direction of the force. The answer is 121.7N which is ok.But the direction is given 34.7degrees.i am getting 25.28.The angle is tan theta=80sin60/60+80cos60 or 60sin60/80+60cos60. I feel it is the latter. But in my book it is80sin60/...... Pls help me.Thanks and regards,Jai

bharat bajaj IIT Delhi
askIITians Faculty 122 Points
9 years ago
Hi, your answer is correct and the book one’s too. The difference between the answers is their representation. You are saying 34.7 degrees with the 80 N force and the book says 25.28 with the 60 N force. These angles sum up to 60 degree. Hence, both are same.
Thanks & Regards
Bharat Bajaj
IIT Delhi
Sumit Majumdar IIT Delhi
askIITians Faculty 137 Points
9 years ago
Dear student,
The resultant force would be given by:
$R=\sqrt{80^{2}+60^{2}+2\left ( 80 \right )\left ( 60 \right )cos\left ( 60^{0} \right )}=\sqrt{6400+3600+2\times 80\times 60\times \frac{1}{2}}=\sqrt{14800}=121.655 N$
The direction would be given by:
Now, lets place the 60 N force along the horizontal direction so the 80 N force would be making an angle of 60 degree.
Hence if we say that the resultant force is making an angle $\alpha$ with the horizontal, we would get:
$Fcos\alpha =60+80cos60; Fsin\alpha=80 sin60$Taking a ratio, would give the required result.
Regards
Sumit
Ankit Jaiswal
165 Points
8 years ago

The resultant force would be given by:
$R=\sqrt{80^{2}+60^{2}+2\left ( 80 \right )\left ( 60 \right )cos\left ( 60^{0} \right )}=\sqrt{6400+3600+2\times 80\times 60\times \frac{1}{2}}=\sqrt{14800}=121.655 N$
The direction would be given by:
Now, lets place the 60 N force along the horizontal direction so the 80 N force would be making an angle of 60 degree.
Hence if we say that the resultant force is making an angle $\alpha$ with the horizontal, we would get:
$Fcos\alpha =60+80cos60; Fsin\alpha=80 sin60$Taking a ratio, would give the required result.