A box of mass M = 10 Kg rests on a 35° inclined plane with the horizontal. A string is used to keep the box in equilibrium. The string makes an angle of 25 ° with the inclined plane. The coefficient of friction between the box and the inclined plane is 0.3.
Answers
Answer:
what is your question first m=10kg
Answer:
The coefficient of friction between the box and the inclined plane is 0.3., T = 52.58 N
Explanation:
From the above question,
They have given :
The tension in the string is equal to the normal reaction force between the box and the inclined plane multiplied by the sine of the angle between them (25°).
T = Nsin(25°)
Therefore,
T = (Mgsin(35°))sin(25°)
The tension in the string required to keep the box of mass M = 10 Kg in equilibrium on a 35° inclined plane is 52.58 N.
This is calculated by multiplying the normal reaction force between the box and the inclined plane, which is equal to Mgsin(35°), by the sine of the angle between the string and the inclined plane (25°).
T = (10 x 9.81 x sin(35°) ) sin (25°)
T = 52.58 N
The coefficient of friction between the box and the inclined plane is 0.3. This tension must be greater than the force of friction acting on the box in order to keep it in equilibrium. If the tension is not greater than this, the box will slide down the inclined plane.
For more such related questions : https://brainly.in/question/22076267
#SPJ3