A 70kg object is being pulled up a slope of 30 degrees such that the

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SUMMARY

The tension in the rope required to pull a 70kg object up a 30-degree slope with a coefficient of friction of 0.3 is calculated using the formula T = mg sin(θ) + f, where f is the frictional force. The gravitational force component along the slope is 343N, and the frictional force can be determined using f = μN, where N is the normal force. The normal force is calculated as N = mg cos(θ), leading to a total tension of 392.1N when friction is considered.

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A 70kg object is being pulled up a slope of 30 degrees such that the rope is parallel to the slope. v is constant. coefficient of friction = 0.3.

what is the tension of the rope?



I know without friction its, something like, T=mgsin(30)=70kg(9.8m/s^2sin(30))= 343N
But i just don't no know to use the friction coefficient? Any help would be much appricated...
 
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