What should be the tension in the string with which the block is pulled upwards for it to move.
A block weighing 120 kg is placed on an incline of 60 degrees with a coefficient of static friction 0.89 and pulled upwards with a string.
1 Answer | Add Yours
The block 120 kg is placed on an incline of 60 degrees. The coefficient of static friction between the block and the surface it is placed on is 0.89. When the block is pulled with a string, the tension in the string has to be determined for which the block moves.
The mass of 120 kg can be divided into 2 components, one normal to the inclined surface equal to 120*9.8*cos 60. The other component is parallel to the surface and equal to 120*9.8*sin 60. If the block has to move it has to be pulled by a force equal to the sum of the gravitational force acting downwards as well as the force of static friction.
The force of static friction is 120*9.8*cos 60*0.89. The total force that has to be overcome is 120*9.8*cos 60*0.89 + 120*9.8*sin 60 = 523.32 + 1018.44 = 1541.76 N
The tension in the string has to be greater than 1541.76 N for the block to be pulled upwards.
We’ve answered 317,615 questions. We can answer yours, too.Ask a question